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Allisa [31]
3 years ago
12

Copper has two isotopes. The one isotope (Copper-63) has an abundance of 69.2% and a mass of 62.930 amu. The second isotope (Cop

per-65) has an abundance of 30.8% and a mass of 64.928 amu. Calculate the average atomic mass of copper. 62.930 amu 63.545 amu 63.929 amu 64.928 amu
Chemistry
1 answer:
Step2247 [10]3 years ago
8 0
Multiply the percentage by the mass for both and add them to get 63.545 amu 
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Students are investigating the relationship between pressure and temperature in gases. They have blown up a balloon with carbon
Ierofanga [76]

Answer:

b

Explanation:

8 0
3 years ago
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A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0
3 years ago
Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?
mylen [45]

Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

<u>Explanation:</u>

Equilibrium expression is denoted by Keq.

Keq is  the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

Therefore, Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

4 0
3 years ago
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Answer:

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Ammonia Formula: NH3

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H2SO4 + 2NH3 = 2NH4+ + SO4 2-

H2SO4 + 2NH3 = (NH₄)₂SO₄

H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L

= 0.08 mol L-1

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