All categories

3 years ago

A compound is 87.1% ag & 12.9% s. find the empirical formula

1 answer:

8 0

Empirical formula of the compound is the simplest ratio of components making up the compound.
In 100 g there’s 87.1 g of Ag and 12.9 g of S.
Let’s calculate for 100 g of the compound
Ag S
Mass 87.1 g 12.9 g
Moles. 87.1 /107.8 g/mol 12.9/32 g/mol
=0.8 mol =0.4mol
Divide by the least number of moles
0.8/0.4 =2 0.4/0.4=1
Ratio of Ag to S is 2:1
Therefore empirical formula is Ag2S

You might be interested in

Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

8 0

3 years ago

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f

pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

What would happen to the gas pressure inside the cylinder if you do the following?

Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.

T₂ = 0.5 T₁. According to Gay-Lussac's law,

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}$

Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

$P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}$

P₂ in terms of the ideal gas equation is:

$P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}$

3 0

3 years ago

Plete the following problems significant figures: 6 × 0.30

jeka57 [31]

6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.

0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.

3 0

3 years ago

Sodium carbonate (Na2CO3) reacts with acetic acid (CH3COOH) to form sodium acetate (NaCH3COO), carbon dioxide (CO2), and water (

Evgen [1.6K]

Answer:

=3,723.3 J=3.72 but if it has the option of -3.72 kJ then use that

Explanation:

Use the formula q=m×Cp×delta T

m=1.500 kg=1,500 g

Co=2.52 J/g·k

delta T=0.985k

q=(1,500g)(2.52 J/g·k)(0.985k)

7 0

3 years ago

Please help me with this

Harrizon [31]

I and Ca is the answer

3 0

3 years ago