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Y_Kistochka [10]
3 years ago
8

A 13.00 g sample of a compound contains 4.15 g potassium (k), 3.76 g chlorine (cl), and oxygen (o). calculate the empirical form

ula.
Chemistry
1 answer:
MrRissso [65]3 years ago
7 0

To solve this problem, let us all convert the mass of each element into number of moles using the formula:

moles = mass / molar mass

Where,

molar mass K = 39.10 g / mol

<span>molar mass Cl = 35.45 g / mol</span>

molar mass O = 16 g / mol

<span>and mass O = 13 g – 4.15 g – 3.76 g  = 5.09 g</span>

 

moles K = 4.15 g / (39.10 g / mol) = 0.106 mol

<span>moles Cl = 3.76 g / (35.45 g / mol) = 0.106 mol</span>

moles O = 5.09 g / (16 g / mol) = 0.318 mol

 

The ratio becomes:

0.106 K: 0.106 Cl: 0.318 O

We divide all numbers with the smallest number, in this case 0.106. This becomes:

K: Cl: 3O

 

Therefore the empirical formula is:

KClO_{3}

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Answer:

C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2

Explanation:

Based on combined gas law, gases under the same pressure, temperature and volume have the same number of moles. With this information we can say the rigth statement is:

<h3>C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2</h3>
7 0
3 years ago
Determine the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C. Carry out two c
Nuetrik [128]

The pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.

<h3>How to calculate pressure?</h3>

The pressure of an ideal gas can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • n = number of moles
  • R = gas law constant
  • T = temperature

According to information in this question;

  • T = 25°C = 25 + 273 = 298K
  • V = 244.6mL = 0.24L
  • R = 0.0821 Latm/Kmol

P × 0.24 = 1 × 0.0821 × 298

0.24P = 24.47

P = 24.47/0.24

P = 101.94atm

Therefore, the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.

Learn more about pressure at: brainly.com/question/11464844

3 0
2 years ago
How many moles of hydrogen are in 3. 06 × 10⁻³ g of glycine , c₂h₅no₂?.
Gala2k [10]

Answer:

n = 6.06 x 10^{-4} mol

Explanation:

n =?

m = 3.06 x 10-³ g

M (H5) = 5 x 1.01 (Since we only want hydrogen)

Atomic mass of C = 12.01

Atomic mass of H is 1,01, etc.

Having this data, we can use the Molar mass formula and change it so we can know the quantity of matter (n) in moles, and we just replace it.

M = \frac{m}{n} ⇔ n = \frac{m}{M} ⇔ n = \frac{3.06 x 10^{-3} }{5,05} ⇔ n = 6.06 x 10^{-4} mol

Note: The numbers I've used may be different from yours, by a small difference. I don't know if it's the case, but hope it helped.

8 0
2 years ago
What coefficient values will balance the reaction shown? CS2 + Cl2? CCl4 + S2Cl2 A. 1,1,1,1 B. 1,3,1,1 C. 2,1,1,1 D. 1,2,1,2
Talja [164]
The answer is B. 1,3,1,1.

Brady
5 0
3 years ago
8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]
yKpoI14uk [10]

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡  ΔH_{f} = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8  kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5  kJ/mol

Equation 4 ≡ BE(Cl2) = 226  kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332  kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

4 0
3 years ago
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