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.
A solution of a substance ‘X’ is used for white washing.
Answer:
The substance which is used for white washing is Calcium oxide . Calcium oxide when reacted with water produces Calcium hydroxide(Lime water).
Explanation:
According to the graph, predict the force of gravity on an object of mass 1000 kg if the gravitational force is 25 Newtons for an object of 500 kg.
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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whats the mass of 4.35x 10^-2 mol of NAOH
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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How any kPa is 2L of gas if it takes 2.55L at 146kPa?
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.
Please I need help thank you
Answer:
its sodium hydroxide
Explanation:
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)
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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If 45 Ml of a 6.5 M solution of HNO3 is added to 55 ml of water what is the new concentration?
Answer:
2.93M
Explanation:
C1V1=C2V2
V2 = (45+55)
45 * 6.5 = ( 100)X
make x subject of formula
:. x = 2.93M
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.
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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a chemical symbol is to an element as a chemical formula is to a
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.
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?
why does lead exist in a higher amount in brown algae than plankton?
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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What is the oxidation number of Boron? (2.2.1)
2+
2-
3+
3-
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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calculate the volume of oxygen produced in the decomposition of 5 moles of KCLO3 at stp
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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1 answer
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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.
Calculate the proper number of significant digits, the density of a 23.23g box occupying 26.5 mL.
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
What element is this in the diagram to the right? (2.1.4)
a. Calcium
b. Aluminum
c. Magnesium
d. silicon
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.
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?
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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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)?
Answer:
the answer is 57.03%
Explanation:
%yield= ((practical yield)/(theoretical yield))×100%
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 .
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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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?
I need help with this homework thank you
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.
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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How many moles of atoms are in 150 g S
Answer:
Approximately 4.678 moles
Explanation:
150/32.065 (atomic weight of S)
Answer:
4.677 Moles
Explanation:
150g / 32.07g = 4.677268475 moles
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
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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Metal D Most reactive
Sodium
Magnesium
Carbon
Metal E
Iron
Hydrogen
Copper Least reactive
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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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.
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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Use the following percent compositions to determine the molecular formula for a compound with a molar mass of 425.82 g/mol. P= 43.7% O= 56.3%
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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The molar mass of argon, Ar, is 39.95 g/mol
How many grams of argon are in 3.41 moles of argon
Answer:
136.2295 grams of Ar
Explanation:
Simply multiply the moles by the grams. 3.41x39.95=136.2295.
A beaker of cold water is placed in a hot water bath at 90°C.Explain what happens in terms of heat diffusion.
Explanation: heat is transferred from the hot water to the cold water until they reach the same temperature
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
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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