What is the oxidation number of Boron? (2.2.1)
2+
2-
3+
3-

What Is The Oxidation Number Of Boron? (2.2.1)2+2-3+3-

Answers

Answer 1

Some boron compounds and the determination of boron's oxidation number. In those boron hydrides that contain one or more B-B bonds, the oxidation number of boron can be less than +3 and more than 0.

Thus, The same molecule will have various boron atom types with various oxidation values in such a complex. Therefore, the average oxidation number would be determined using the formula for such a molecule.

Tetraborane (B4 H10) and decaborane (B10 H14) are two examples of such compounds that are displayed in the table's final two entries.

These substances are less stable and have complicated structures. The majority of stable boron compounds have boron with an oxidation number of +3.

Thus, Some boron compounds and the determination of boron's oxidation number. In those boron hydrides that contain one or more B-B bonds, the oxidation number of boron can be less than +3 and more than 0.

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Related Questions

Explain two positive aspects of using methane recapture systems.

Answers

Answer:

Two positive aspects of using methane recapture systems are able to generate significant electricity. Another benefit is that the process of anaerobic digestion creates heat that can be used to warm buildings where animals are kept

Answer:  The correct answer is;

Two positive aspects of using methane recapture systems include lowering the impact on greenhouse gasses and the production of energy. Methane is a very potent greenhouse gas that is contributing to global warming. As a result, the recapturing process reduces the methane impacts of global warming by reclaiming and reusing the gas for other purposes. Recaptured methane can be stored and used to generate electricity or used as fuel to power updated vehicles and other engines on the farm. The overall benefits from this combination are reducing impacts causing global warming and lower the cost of electricity or fuel on the farm.

Explanation:  This answer has been confirmed correct.

why does lead exist in a higher amount in brown algae than plankton?​

Answers

Lead levels in plankton and algae are high, mostly as a result of environmental pollution brought on by human activity. While it is true that some brown algae species have the ability to accumulate heavy metals like lead.

Plankton and algae have high levels of lead, mostly as a result of environmental contamination brought on by human activities including mining, industrial operations, and the burning of fossil fuels.

Due to the fact that plankton and algae take up trace quantities of lead from the surrounding water, their tissues contain greater concentrations of the metal.

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Question 1
Given the equation: Q = mcAT
Q = heat (in Joules)
m = mass (in grams)
C = 4.18 (specific heat capacity)
AT change in temperature (°C)
How many Joules of heat energy are absorbed when 200 grams of water are heated from 20 C to 60 C.

Answers

The amount of heat energy absorbed when 200 grams of water are heated from 20 C to 60 C is 33,440 Joules.

To find the amount of heat energy absorbed when 200 grams of water are heated from 20 C to 60 C, we can use the equation Q = mcAT.
First, we need to find the value of m, which is the mass of the water in grams. In this case, it is given as 200 grams.
Next, we need to find the value of AT, which is the change in temperature in degrees Celsius.

This can be calculated by subtracting the initial temperature from the final temperature, which gives us 60 C - 20 C = 40 C.
The specific heat capacity of water, C, is given as 4.18 Joules per gram per degree Celsius.
Now we can plug in the values into the equation:
Q = mcAT
Q = (200 g) x (4.18 J/g°C) x (40°C)
Q = 33,440 J
Therefore, the amount of heat energy absorbed when 200 grams of water are heated from 20 C to 60 C is 33,440 Joules.

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Please I need help thank you

Answers

Answer:

its sodium hydroxide

Explanation:

The evidence of quantized energy states in atoms comes from
1. photoelectric effect
2. rainbows from prisms
3. oil drop experiment
4. diffraction
5. bright line or emission spectra 6. gold foil experiment

Answers

The evidence for quantized energy states in atoms stems primarily from the photoelectric effect, bright line spectra, and diffraction phenomena. Options 1,4 and 5 are correct.

The evidence of quantized energy states in atoms comes from several experimental observations, which collectively provide a comprehensive understanding of atomic structure and the behavior of electrons within atoms.

One of the key pieces of evidence is the observation of the photoelectric effect (1). When light shines on a metal surface, electrons are ejected from the surface. The observation that electrons are only ejected if the light has a minimum frequency, regardless of its intensity, supports the idea that energy is quantized in discrete packets known as photons.

Another crucial observation is the presence of bright line or emission spectra (5). When atoms are excited, they emit light at specific wavelengths that correspond to distinct energy transitions. These discrete emission lines indicate that electrons can only exist in specific energy levels within an atom, and they transition between these levels by absorbing or emitting photons of precise energy.

The phenomenon of diffraction (4) also provides evidence for quantized energy states. Diffraction occurs when light passes through a narrow slit or encounters a periodic structure. The resulting pattern indicates that light behaves as waves with specific wavelengths. This suggests that the energy of light is quantized and can only exist in certain discrete values.

While rainbows from prisms (2) and the oil drop experiment (3) are not directly related to quantized energy states in atoms, they are important experiments in their own right. Rainbows result from the dispersion of white light into its component colors due to different wavelengths of light bending at different angles. The oil drop experiment explores the behavior of charged oil droplets in an electric field, providing insights into charge quantization.

Lastly, the gold foil experiment, also known as the Rutherford scattering experiment, is significant but not directly related to quantized energy states. It demonstrated that the atom has a small, dense, positively charged nucleus by observing the deflection of alpha particles fired at a thin gold foil.

These experimental observations support the fundamental concept that energy levels in atoms are discrete and that electrons occupy specific energy states within an atom.  Options 1,4 and 5 are correct.

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Choose two regions to compare the effects of climate change in areas. Comment on things like major events, adaptation, the carbon cycle and the effect on humans.

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The two regions  that I will  compare their effects of climate change in areas are Arctic and the Amazon rainforest..

What is the comparism?

The Major Events that can be associated to Arctic region  can be described as rapid warming that affect ecosystem.

The major  that can be attributed to Amazon rainforest  can be described as increased deforestation rates.

In term of Adaptation the Arctic communities  are facing some challenges which makes some of the people to communities relocating homes away from eroding coastline.

In term of Adaptation the Amazon rainforest were seeking for the way to combat deforestation  and bring about Initiatives such as reforestation.

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Using the information in the table to the right, calculate the average atomic mass of strontium. Report to two decimal places.
A 3-column table with 4 rows titled Strontium. Column 1 is labeled Isotope with entries upper S 4 84, upper S r 86, upper S r 87, upper S r 88. Column 2 is labeled Mass in atomic mass units with entries 83.913428, 85.909273, 86.908902, 87.905625. Column 3 is labeled abundance with entries 0.56 percent, 9.86 percent, 7.00 percent, 82.58 percent.

Answers

The column 1 has the value of Isotope, column 2 has the value of mass in atomic mass units, and column 3 has the value of abundance and the average atomic mass of strontium is 87.47 amu.

To calculate the average atomic mass of strontium using the given information, we need to multiply the mass of each isotope by its abundance and then sum up these values. Here's the calculation:

Isotope | Mass (amu) | Abundance

^84Sr | 83.913428 | 0.56%

^86Sr | 85.909273 | 9.86%

^87Sr | 86.908902 | 7.00%

^88Sr | 87.905625 | 82.58%

To find the average atomic mass, we multiply each isotope's mass by its abundance (in decimal form) and sum up the values:

Average atomic mass = ([tex]Mass of ^{84Sr}[/tex] × [tex]Abundance of^{84Sr}[/tex]) + ([tex]Mass of ^{86Sr}[/tex]× [tex]Abundance of^{86Sr}[/tex]) + ([tex]Mass of ^{87Sr}[/tex] × [tex]Abundance of^{87Sr}[/tex]) + ([tex]Mass of ^{88Sr}[/tex] × [tex]Abundance of^{88Sr}[/tex])

Average atomic mass = (83.913428 amu × 0.0056) + (85.909273 amu × 0.0986) + (86.908902 amu × 0.0700) + (87.905625 amu × 0.8258)

Calculating this expression yields:

Average atomic mass = 0.469901638 + 8.468098826 + 6.08462314 + 72.44409075

= 87.466714354 amu

Rounding the result to two decimal places, the average atomic mass of strontium is approximately 87.47 amu.

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I need help with this homework thank you

Answers

S+H2SO4 arrow sign H2O+ SO2 that is the answer

Metal D Most reactive
Sodium
Magnesium
Carbon
Metal E
Iron
Hydrogen
Copper Least reactive

Answers

As per the given details, Metal D is extracted from its oxide by reduction with hydrogen, and Metal E is removed from the earth as the metal itself.

Based on the provided information, we can match the metals to the methods used to extract them as follows:

Sodium - Extracted by electrolysis of a molten ionic compound.

Magnesium - Extracted from its oxide by reduction with carbon.

Carbon - Not a metal, so it doesn't apply in this context.

Metal E - Extracted from its oxide by reduction with hydrogen.

Iron - Removed from earth as metal itself.

Hydrogen - Not a metal, so it doesn't apply in this context.

Copper - Not a metal D or E, so it doesn't apply in this context.

Matching the metals to the extraction methods:

Sodium - extracted by electrolysis of a molten ionic compound.

Magnesium - extracted from its oxide by reduction with carbon.

Metal D - extracted from its oxide by reduction with hydrogen.

Metal E - removed from earth as metal itself.

Iron - removed from earth as metal itself.

Therefore, Metal D is extracted from its oxide by reduction with hydrogen, and Metal E is removed from the earth as the metal itself.

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calculate the volume of oxygen produced in the decomposition of 5 moles of KCLO3 at stp

Answers

The decomposition of potassium chlorate KClO₃ in the presence of manganese oxide MnO is given by the reaction equation:

KClO₃ (s) → 2KCl (s) + 3O₂ (g)

To calculate the moles of product formed from moles of reactants, the following steps are followed:

1. Balancing the equation

2. Calculating the ratio of product's stoichiometric coefficient and reactant's stoichiometric coefficient.

3. Multiplying the obtained ratio with the number of moles of reactant.

Thus, the number of moles of oxygen evolved will be calculated as:

R = [tex]\frac{coefficient of O2}{coefficient of KClO3}[/tex] = [tex]\frac{3}{1}[/tex] = 3

Number of moles of oxygen evolved = R × number of moles of KClO₃ = 3×5= 15 moles

From the ideal gas equation, 1 mole of gas is equivalent to 22400 ml or 22.4 L.

Thus, volume of oxygen evolved = 22400 × 15 = 336000 ml = 336 L

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When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl, what is the percentage yield, according to ch4(g) + cl2(g) → ch3cl(g) + hcl(g)?

Answers

Answer:

the answer is 57.03%

Explanation:

%yield= ((practical yield)/(theoretical yield))×100%

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions.

1)Ni+(aq)→Ni2+(aq)+Ni(s) (acidic solution)
2)MnO42−(aq)→MnO4−(aq)+MnO2(s) (acidicsolution)
3)H2SO3(aq)→S(s)+HSO−4(aq) (acidicsolution)
4) Cl2(aq)→Cl−(aq)+ClO−(aq) (basicsolution)
Express your answer as a chemical equation including phases.

Answers

The balance reactions are :

1) 2Ni+(aq) + 4H+(aq) → 2Ni2+(aq) + 2Ni(s) + 2H2O(l) ,

2) 3MnO42-(aq) + 4H+(aq) → 2MnO4-(aq) + MnO2(s) + 2H2O(l) ,

3) H2SO3(aq) + H2O(l) → S(s) + 2HSO4-(aq) + 2H+(aq) ,

4) Cl2(aq) + 2OH-(aq) → Cl-(aq) + ClO-(aq) + H2O(l).

1) To balance the disproportionation reaction of Ni+ in an acidic solution, we need to ensure that the number of atoms and charges are balanced on both sides of the equation. The balanced equation is as follows:

2Ni+(aq) + 4H+(aq) → 2Ni2+(aq) + 2Ni(s) + 2H2O(l)

In this reaction, Ni+ is oxidized to Ni2+ (oxidation state increases from +1 to +2) while simultaneously being reduced to Ni (oxidation state decreases from +1 to 0). The hydrogen ions (H+) act as the oxidizing agent, accepting electrons and being reduced to form water (H2O).

2) Balancing the disproportionation reaction of MnO42- in an acidic solution:

3MnO42-(aq) + 4H+(aq) → 2MnO4-(aq) + MnO2(s) + 2H2O(l)

In this reaction, MnO42- is both oxidized and reduced. The oxidation state of Mn in MnO42- changes from +7 to +6 in MnO4- (reduction) and from +7 to +4 in MnO2 (oxidation). The hydrogen ions (H+) again act as the oxidizing agent, undergoing reduction to form water.

3) Balancing the disproportionation reaction of H2SO3 in an acidic solution:

H2SO3(aq) + H2O(l) → S(s) + 2HSO4-(aq) + 2H+(aq)

In this reaction, H2SO3 is both oxidized and reduced. The sulfur (S) in H2SO3 is reduced from an oxidation state of +4 to 0 in S, while the hydrogen sulfite ion (HSO3-) is oxidized from an oxidation state of +4 to +6 in HSO4-. The water molecule (H2O) acts as a reactant and is not involved in the redox process.

4) Balancing the disproportionation reaction of Cl2 in a basic solution:

Cl2(aq) + 2OH-(aq) → Cl-(aq) + ClO-(aq) + H2O(l)

In this reaction, Cl2 is both oxidized and reduced. The chlorine (Cl) in Cl2 is reduced from an oxidation state of 0 to -1 in Cl-, while simultaneously being oxidized from an oxidation state of 0 to +1 in ClO-. The hydroxide ions (OH-) act as the reducing agent, accepting electrons and being oxidized to form water (H2O). The reaction takes place in a basic solution, hence the presence of OH- ions.

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1 answer

......................................................

Answers

The most accurate statement about signal transmissions among the given options is:

a) All signals in transmission will lose clarity with distance.

When a signal is transmitted over a distance, it can experience various types of degradation or attenuation. Factors such as distance, interference, noise, and the medium through which the signal travels can all contribute to a reduction in the clarity or quality of the signal. This means that as the distance between the source and receiver increases, the signal may become weaker, distorted, or prone to interference, resulting in a loss of clarity.

a chemical symbol is to an element as a chemical formula is to a

Answers

A chemical symbol is to an element as a chemical formula is to a **compound**.

A chemical symbol is a one- or two-letter designation of an element. For example, the symbol for oxygen is O. A chemical formula is a combination of chemical symbols that shows the elements in a compound and the relative proportions of those elements. For example, the chemical formula for water is H2O, which means that water is made up of two hydrogen atoms and one oxygen atom.

So, a chemical symbol is a short way of representing an element, while a chemical formula is a short way of representing a compound.

How many ATOMS of sulfur are present in 5.21 grams of sulfur dichloride? ________atoms of sulfur

How many GRAMS of chlorine are present in 5.86 * 10 raised 22 molecules of sulfur dichloride?

__________grams of chlorine .

Answers

0.0506 moles or 0.0506 x 6.02 x 10^23 atoms of sulfur is present in 5.21 grams of sulfur dichloride and 6.89 grams of chlorine is present in 5.86 x 10^22 molecules of sulfur dichloride.

Thus, the molar mass of Sulphur dichloride should be known in order to calculate the quantity of Sulphur atoms in 5.21 grammes of Sulphur dichloride. The molar mass of sulphur dichloride is about 102.97 g/mol.

We may determine how many moles of Sulphur dichloride there are in 5.21 grammes using the molar mass: Mass / Molar mass = number of moles Therefore, 0.0506 moles are equal to 5.21 g divided by 102.97 g/mol.

The molar mass of chlorine, which is roughly 35.45 g/mol, may be multiplied by the number of moles to determine the mass: Chlorine mass is equal to 6.89 grammes (0.1944 mol x 35.45 g/mol) n 5.86 x 10^22 molecules of sulfur dichloride.

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why does lead exist in such high concentrations in plankton and algae?​

Answers

[tex]\huge\mathcal{\fcolorbox{aqua}{azure}{\red{Answer:-}}}[/tex]

Lead exists in high concentrations in plankton and algae primarily due to environmental pollution from human activities, such as industrial processes, mining, and the burning of fossil fuels. Plankton and algae accumulate trace amounts of lead from their surrounding water, resulting in higher concentrations within their tissues.

An amateur entomologist captures a particularly excellent ladybug specimen in a plastic jar. The internal volume of the jar is 0.5L, and the air within the jar is initially at 1 atın. The bug-lover is so excited by the catch that he squeezes the jar fervently in his sweaty palm, compressing it such that the final pressure within the jar is 1.25 atm. What is the final volume of the ladybug's prison? ​

Answers

The final volume of the ladybug's prison is approximately 0.4 liters.

To determine the final volume of the ladybug's prison, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. The equation for Boyle's Law is:

P1 * V1 = P2 * V2

Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively.

In this scenario, the initial volume (V1) is given as 0.5 L, and the initial pressure (P1) is 1 atm. The final pressure (P2) is 1.25 atm. We need to find the final volume (V2).

Plugging the given values into the equation, we have:

1 atm * 0.5 L = 1.25 atm * V2

Simplifying the equation, we find:

0.5 L = 1.25 atm * V2

Dividing both sides of the equation by 1.25 atm, we get:

0.5 L / 1.25 atm = V2

V2 ≈ 0.4 L

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a 250 ml flask of hydrogen gas is collected at 763 mmHg and 35C by displacement of water from the flask. the vapor pressure of water at 35c is 42.2 mmhg. how many moles of hydrogen gas are in the flask? (think ideal gas law and dalton's law of partial pressure)

Answers

There are approximately 0.0112 moles of hydrogen gas in the 250 ml flask.

To determine the number of moles of hydrogen gas in the flask, we can use the ideal gas law and Dalton's law of partial pressure.

First, let's convert the given pressures to atm units:

P_total = P_hydrogen + P_water vapor

P_total = (763 mmHg - 42.2 mmHg) / (760 mmHg/atm) [Converting to atm]

P_total = 0.9524 atm

Next, let's convert the given temperature to Kelvin:

T = 35°C + 273.15 [Converting to Kelvin]

T = 308.15 K

Now we can use the ideal gas law equation: PV = nRT

R is the ideal gas constant, which has a value of 0.0821 L·atm/(mol·K)

Rearranging the equation to solve for n (moles):

n = PV / RT

Substituting the given values:

n = (0.9524 atm) * (0.250 L) / (0.0821 L·atm/(mol·K)) * (308.15 K)

Simplifying the expression:

n = 0.0112 mol

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Calculate the proper number of significant digits, the density of a 23.23g box occupying 26.5 mL.

Answers

Answer:

0.877 mL

Explanation:

The box's density would be the ratio of the mass of the box and its volume
which is, (23.23/26.5) mL
or, 0.8766 mL
We must round this down to 3 significant figures,
which will be 0.877 mL

According to the following reaction, how many moles of hydrobromic acid are necessary to form 0.723 moles bromine?
2HBr(aq) → H₂(g) + Br₂(l)
How many mol of hydrobromic acid?

Answers

To answer this question you need to use stoichiometry

We have the number of moles for bromine, we can use that, and the equation is balanced so we can move on

If you look at the reaction, how many moles of HBr are there per 1 Br2? There’s 2/1 or just 2 HBr per 1 Br2. Multiply that with 0.732
2 x 0.732 = 1.464 mol
Hope this helps

How many moles of atoms are in 150 g S

Answers

Answer:

Approximately 4.678 moles

Explanation:

150/32.065 (atomic weight of S)

Answer:

4.677 Moles

Explanation:

150g / 32.07g = 4.677268475 moles

A solution of a substance ‘X’ is used for white washing.

Answers

Answer:

(If you like this answer i would appreciate if u give brainliest but otherwise, i hope this helped ^^)

Explanation:

White washing is a traditional technique where a mixture or solution is applied to surfaces to give them a white appearance. The substance 'X' mentioned in your question could refer to various materials or chemicals commonly used in white washing. Some common substances used for white washing include lime, chalk, or a combination of lime and water.

Lime is a key component in many white washing solutions. It is derived from heating limestone or chalk, which produces calcium oxide (quicklime). Quicklime is then slaked with water to produce calcium hydroxide (slaked lime). The slaked lime is mixed with water to form a white wash solution.

Chalk, when ground into a fine powder, can also be used as a whitening agent in white wash solutions. The chalk particles are mixed with water to form a paste or solution.

Both lime and chalk-based white wash solutions provide a thin, breathable coating that adheres to surfaces and helps protect them while giving a white appearance. The solution can be applied to various surfaces, including walls, fences, and even trees or structures in the outdoors.

It's important to note that the specific recipe for white wash solutions may vary depending on regional preferences and desired effects. Additionally, the application techniques and preparations may differ based on the surface being treated.

According to the following reaction, how many moles of phosphoric acid will be formed upon the complete reaction of
0.949 moles perchloric acid (HCIO4) with excess tetraphosphorus decaoxide?
12HClO4 (aq) + P4O10 (s)→ 4H3PO4 (aq) + 6C1₂O7(l)
How many moles of phosphoric acid?

Answers

4 moles of phosphoric acid will be formed upon the complete reaction of 0.949 moles perchloric acid with excess tetraphosphorus decaoxide.

Use the following pairs of standard reduction potentials below to answer question 21. Respectively is A. Is the Half-reaction, and B. Is E^0(volts) ; A. Cr^ 2+ +2e^- ->Cr B.-0.913 A. Fe^2+ +2e^->Fe B. -0.447 A. Cd^2+ +2e^-> Cd B.-0.4030 A.Br2+2e^-> 2Br B. +1.06
Question 21. For each of these pairs of half -reactions , write a balanced equation for the overall cell reaction and calculate the standard cell potential, E^0cell A. Half-reactions: Cd^ 2+ (aq)+2e^ -> Cd(s); Cr^ 2+ (aq)+2e^-> Cr(s) Cell reaction : E^0cell: B. Half-reactions: Fe^ 2+ (aq)+2e^ -> Fe(s); Br2 (g)+2e^- ->2Br^ - (aq) Cell reaction : E^0cell

Answers

A. Cell reaction:[tex]2 Cd^2+(aq) + 2 Cr(s) - > 2 Cd(s) + 2 Cr^2+(aq) ; E^0cell = -0.51 V[/tex]

B. Cell reaction: [tex]2 Fe^2+(aq) + Br2(g) - > 2 Fe(s) + 2 Br^-(aq) ; E^0cell = +1.507[/tex]

Let's calculate the standard cell potential, E^0cell, for each pair of half-reactions and write the balanced equations for the overall cell reactions:

A. Half-reactions:

[tex]Cd^2+(aq) + 2e^- - > Cd(s) (E^0 = -0.403 V)\\Cr^2+(aq) + 2e^- - > Cr(s) (E^0 = -0.913 V)[/tex]

To calculate the standard cell potential, we subtract the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction).

[tex]E^0cell = E^0cathode - E^0anode\\E^0cell = (-0.913 V) - (-0.403 V) = -0.51 V[/tex]

The balanced equation for the overall cell reaction is obtained by multiplying the half-reactions by coefficients to ensure that the number of electrons transferred is the same:

[tex]2 Cd^2+(aq) + 2 Cr(s) - > 2 Cd(s) + 2 Cr^2+(aq)[/tex]

B. Half-reactions:

[tex]Fe^2+(aq) + 2e^- - > Fe(s) (E^0 = -0.447 V)\\Br2(g) + 2e^- - > 2 Br^-(aq) (E^0 = +1.06 V)\\E^0cell = E^0cathode - E^0anode\\E^0cell = (+1.06 V) - (-0.447 V) = +1.507[/tex]V

The balanced equation for the overall cell reaction is:

[tex]2 Fe^2+(aq) + Br2(g) - > 2 Fe(s) + 2 Br^-(aq)[/tex]

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A compound is found to contain 3.622 % carbon and 96.38 % bromine by mass.

To answer the question, enter the elements in the order presented above.

QUESTION 1:
The empirical formula for this compound is
.


QUESTION 2:
The molecular weight for this compound is 331.6 amu.

The molecular formula for this compound is

Answers

Question 1 : The empirical formula for this compound is CBr₄.

Question 2: The molecular formula of the compound is CBr₄.

To determine the empirical formula of the compound, we need to find the simplest whole-number ratio between the elements present.

Empirical formula:

The compound contains 3.622% carbon and 96.38% bromine. To convert these percentages into masses, we can assume a 100 g sample of the compound.

Mass of carbon = (3.622/100) * 100 g = 3.622 g

Mass of bromine = (96.38/100) * 100 g = 96.38 g

Next, we need to find the moles of each element. We can use their atomic masses to convert the masses to moles.

Atomic mass of carbon (C) = 12.01 g/mol

Atomic mass of bromine (Br) = 79.90 g/mol

Moles of carbon = Mass of carbon / Atomic mass of carbon = 3.622 g / 12.01 g/mol ≈ 0.3017 mol

Moles of bromine = Mass of bromine / Atomic mass of bromine = 96.38 g / 79.90 g/mol ≈ 1.205 mol

To find the simplest whole-number ratio between the elements, we divide both moles by the smallest number of moles (0.3017 mol in this case):

Moles of carbon (C) = 0.3017 mol / 0.3017 mol = 1

Moles of bromine (Br) = 1.205 mol / 0.3017 mol ≈ 4

Therefore, the empirical formula of the compound is CBr₄.

Molecular formula:

The empirical formula of CBr₄ gives us the simplest whole-number ratio of the elements. To determine the molecular formula, we need the molar mass of the compound.

Given that the molecular weight (molar mass) of the compound is 331.6 amu, we can find the ratio of the molecular weight to the empirical formula weight:

Molecular weight / Empirical formula weight = 331.6 amu / (12.01 amu + (4 × 79.90 amu)) ≈ 331.6 amu / 332.64 amu ≈ 0.9965

Since the ratio is close to 1, the empirical formula is also the molecular formula. Therefore, the molecular formula of the compound is CBr₄.

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Categorize the following according to where they should be in the net-ionic equation. The options will not show coefficients. You will not use all options. The net-ionic equation for the reaction of strontium chloride (SrCl₂) and mercury(I) nitrate (Hg2(NO3)2) contains which of the 1. following species? + Sr (NO₂) (NO3)₂ Hg₂012 (5) 201 Possible answers Sr2+ SrC12 Sr Cl₂ + carry Br- Hg₂ 2 CI^- + 2 can Sv Hg2C12 Product(s) + 2 Hg2^2+ Hg₂c1z + Hg2(NO3)2 Sr(NO3)2​

Answers

The species that should be present in the net-ionic equation are:

[tex]Hg_{2}, 2 CI^-[/tex]and [tex]Hg_{2}Cl_{2}[/tex]

To determine the species that should be present in the net-ionic equation for the reaction of strontium chloride (SrCl₂) and mercury(I) nitrate (Hg2(NO3)2), let's analyze the reactants and products:

Reactants:

Strontium chloride (SrCl₂): Sr2+, Cl-

Mercury(I) nitrate (Hg2(NO3)2): Hg2^2+, NO3-

Products:

Strontium nitrate (Sr(NO3)2): Sr2+, NO3-

Mercury(I) chloride (Hg2Cl2): Hg2^2+, Cl-

The species that should be present in the net-ionic equation are:

[tex]Hg_{2}, 2 CI^-[/tex]and [tex]Hg_{2}Cl_{2}[/tex]

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number of SO2 molecules in 1.28moles of SO2

Answers

Answer: There are approximately 7.70176 × 10²³molecules of SO₂ in 1.28 moles of SO₂.

Explanation:

To calculate the number of SO₂ molecules in 1.28 moles of SO₂, we need to use Avogadro's number

Avogadro's number =  6.022 × 10²³ molecules per mole.

First, we'll calculate the number of molecules in 1 mole of SO2:

1 mole of SO₂ = 6.022 × 10²³ molecules.

Then, we'll multiply this value by the number of moles of SO₂ given (1.28 moles):

Number of molecules = 1.28 moles × (6.022 × 10²³ molecules/mole)

Calculating this, we get:

Number of molecules = 7.70176 × 10²³molecules

What element is this in the diagram to the right? (2.1.4)
a. Calcium
b. Aluminum
c. Magnesium
d. silicon

Answers

Answer: b. Aluminum

Explanation:

Frist count all the electrons in the given model. You will get there is 13 electrons. The number of electrons in an element is equivalent to the number of protons in an element. Using a periodic table look for the element that has the equivalent amount of protons. You find that Aluminum has 13 protons, so it is the element shown in the diagram.

A particular natural gas consists, in mole percents, of 83.0 % CH4 (methane), 11.2 % C2H6 (ethane), and 5.80 % C3H8 (propane). A 385- L sample of this gas, measured at 23 ∘C and 739 mmHg, is burned in an excess of oxygen gas. How much heat is evolved in this combustion reaction?

Answers

The heat evolved in the combustion reaction of a 385 L sample of natural gas consisting of 83.0% CH4, 11.2% C2H6, and 5.80% C3H8, measured at 23°C and 739 mmHg, is calculated to be -4.45 x 10^6 kJ.

help please match the items ​

Answers

Correct answers are:

i. Coal, charcoal, oil, and gas - E. Fuels

ii. It supports combustion - D. Fire triangle

iii. Fire associated with electrical equipment - G. Class E fire

iv. A chemical change occurring in iron or steel - C. Rusting

v. Oxygen, heat, and fuel - D. Fire triangle

vi. Fire involving flammable liquids - G. Class B fire

vii. Coating of iron and steel with Zinc - L. Galvanizing

viii. Monoammonium phosphate with Nitrogen carrier - M. Fire extinguisher

ix. A team which put off fire when it's out of control - J. Fire squad

x. It uses oxygen when burning but produces soot - N. Non-luminous flame

Coal, charcoal, oil, and gas are commonly used as fuels. Therefore, they are matched with E. Fuels. Electrical fires are classified as Class E fires. Therefore, it is matched with G. Class E fire. Fires that involve flammable liquids are classified as Class B fires. Therefore, it is matched with G. Class B fire.

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