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

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.

The word equations tells us that the combination of copper oxide and sulfuric acid produces copper sulfate and water

A word equation is a way to represent a chemical reaction using words instead of chemical formulas or symbols. It describes the reactants and products of the reaction in a clear and understandable manner as we see in the question that was shown as equation.

Note that the copper oxide and the sulfuric acids are the reactants that are combined and the copper sulfate and the water are the products.

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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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The mass of 4.35 × 10^-2 mol of NaOH is 1.74 grams.

Given

Number of moles = 4.35 x 10^-2

First, we calculate the molar mass of NaOH,

Molar mass of NaOH = (1 × atomic mass of Na) + (1 × atomic mass of O) + (1 × atomic mass of H)

= (1 × 22.99 g/mol) + (1 × 16.00 g/mol) + (1 × 1.01 g/mol) = 40.00 g/mol

Molar mass of NaOH = 40.00 g/mol

Mass of NaOH = Number of moles × Molar mass

                          = 4.35 × 10^-2 mol × 40.00 g/mol  

Mass of NaOH  = 1.74 g

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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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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.

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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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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1) Diesel has a higher viscosity than petrol.

2) Petrol is more flammable than diesel.

3) The formula will be C₁₀H₂₂.

4) The equation is; 2C8H18+25O2→16CO2+18H2O.

Depending on the precise composition and temperature, the viscosity of gasoline and diesel can change. In general, diesel is more viscous than gasoline. Higher viscosity fluids are thicker and flow more slowly than lower viscosity fluids because viscosity relates to the resistance of a fluid to flow.

Diesel is less flammable than gasoline. The lowest temperature at which gasoline can evaporate and turn into an ignitable combination in air is known as its flash point, and it is lower for gasoline. Compared to diesel fuel, petrol vapors are much more flammable and can ignite at lower temperatures.

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Explanation: heat is transferred from the hot water to the cold water until they reach the same temperature

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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The molecular formula of the compound is P₂O₅.

To determine the molecular formula for a compound with the given percent compositions, we can assume a certain mass for the compound and calculate the number of moles for each element based on its percentage. Then, we can determine the simplest whole-number ratio of the elements to find the molecular formula.

Let's assume a mass of 100 grams for the compound. From the given percent compositions, we have 43.7 grams of phosphorus (P) and 56.3 grams of oxygen (O).

Next, we calculate the moles of each element using their molar masses. The molar mass of phosphorus (P) is 30.97 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.

The moles of phosphorus can be calculated as:

moles of P = mass of P / molar mass of P = 43.7 g / 30.97 g/mol ≈ 1.41 mol

The moles of oxygen can be calculated as:

moles of O = mass of O / molar mass of O = 56.3 g / 16.00 g/mol ≈ 3.52 mol

Now, we find the simplest whole-number ratio of the elements by dividing the moles of each element by the smallest number of moles (1.41 mol in this case):

P:O ≈ 1.41 mol / 1.41 mol : 3.52 mol / 1.41 mol ≈ 1:2.5

Since we need whole-number ratios, we can multiply both numbers by 2 to get a whole number for oxygen:

P:O ≈ 2:5

Therefore, the molecular formula of the compound is P₂O₅.

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Answer:

2.93M

Explanation:

C1V1=C2V2

V2 = (45+55)

45 * 6.5 = ( 100)X

make x subject of formula

:. x = 2.93M

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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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.

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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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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Answer:

the answer is 57.03%

Explanation:

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

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

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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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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Answer:

its sodium hydroxide

Explanation:

Answer:

The substance which is used for white washing is Calcium oxide . Calcium oxide when reacted with water produces Calcium hydroxide(Lime water).

Explanation:

Based on the proportionality established by the given data, we predict that the force of gravity on an object of mass 1000 kg would be 50 Newtons.

To predict the force of gravity on an object of mass 1000 kg, we can use the concept of proportionality. Based on the given information, we have two data points: the mass of an object of 500 kg and the corresponding gravitational force of 25 Newtons.

Let's denote the force of gravity on the object of mass 1000 kg as F₁ and solve for it using a proportion:

mass₁ / force₁ = mass₂ / force₂

Plugging in the values we have:

1000 kg / F₁ = 500 kg / 25 N

Cross-multiplying and solving for F₁:

(1000 kg) * (25 N) = (500 kg) * F₁

25,000 kg*N = 500 kg * F₁

Dividing both sides by 500 kg:

F₁ = (25,000 kg*N) / 500 kg

F₁ = 50 N

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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.

Answer: it will have a pressure of approximately 371.73 κPa.

We know that According to Boyle's Law

P1 ×V1 = P2 ×V2

therefore in order to find P2

P2 =[tex]\frac{P1 X V1}{V2}[/tex] --------------------(I)

where

P1 = initial pressure

V1 = initial volume    

P2 = final pressure

V2 = final volume

here

P1(initial pressure) = 146 kPa        

V1(initial volume) = 2.55 L

V2(final volume) = 2 L

Now putting the given values in equation (I) :

P2 = (146 kPa × 2.55 L) / 2 L

P2 = 371.73 kPa

Therefore, 2 L of gas, when it takes 2.55 L at 146 kPa, will have a pressure of 371.73 kPa.

Answer:

136.2295 grams of Ar

Explanation:

Simply multiply the moles by the grams. 3.41x39.95=136.2295.