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Nostrana [21]
3 years ago
10

In nature, oxygen has three common isotopes. The atomic masses and relative abundances of these isotopes are given in the table

below.
Isotope Atomic Mass (amu) Relative Abundance
O-16          15.995                          99.759%
O-17          16.995                             0.037%
O-18          17.999                               0.204%
Calculate the average atomic mass of oxygen. Show all of your calculations below.
Chemistry
1 answer:
Reil [10]3 years ago
7 0
<span>15.995* 0.99759 + 16.995*0.00037 + 17.999*0.00204 = 15.999</span>
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Given: logK=nE∘0.0592 What is the value K for this redox reaction? Zn2+(aq) + 2 Cl−(aq) → Zn(s) + Cl2(
anygoal [31]

<em>K</em> = 2.4 × 10^(-72)

<em>Step 1</em>. Determine the <em>value of n </em>

Zn^(2+) + 2e^(-) → Zn

2Cl^(-) → Cl_2 + 2e^(-)

Zn^(2+) + 2Cl^(-) → Zn + Cl_2

∴ <em>n</em> = 2

<em>Step 2</em>. Calculate <em>K</em>

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4 0
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I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19
IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

For the 4.12 g sample

Moles of a substance is given by

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}

Number of molecules is given by

nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}

For the 19.37 g sample

n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}

Number of molecules is given by

nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}

1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

7 0
3 years ago
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