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3 years ago

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An unknown element, X, has an atomic mass of 107.868 amu. The X-109 isotope (108.905 amu) is 48.16%. What is the amu of the othe

r isotope (report final answer to the correct number of significant figures)

1 answer:

juin [17]3 years ago

4 0

Answer:

106.905 amu is the mass of the other isotope

Explanation:

The atomic mass of an element is the sum of the masses of the isotopes multiplied by its abundance. The atomic mass of an element X with 2 isotopes is:

X = X-109*i + X-107*i

Where X is the atomic mass = 107.868 amu

X-109 = 108.905amu, i = 48.16% = 0.4816

X-107 = ?, i = 1-0.4816 = 0.5184

Replacing:

107.868amu = 108.905amu*0.4816 + X-107*0.5184

55.4194 = X-107*0.5184

106.905 = X-107

<h3>106.905 amu is the mass of the other isotope</h3>

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11. A 4.175 gram sample of a certain hydrate of copper (II) sulfate, CuSO,• xH,O, is heated until all

Wewaii [24]

The formula of the hydrate = CuSO₄• 3H₂O

<h3>Further explanation</h3>

Given

4.175 grams sample CuSO₄• xH₂O

3.120 grams anhydrous compound CuSO₄

Required

The formula

Solution

mass of H₂O driven off :

= 4.175 - 3.12

= 1.055 g

MW CuSO₄ = 159.5 g/mol

MW H₂O = 18 g/mol

mol ratio of CuSO₄ : H₂O :

= 3.12/159.5 : 1.055/18

= 0.01956 : 0.05861

= 1 : 3

3 0

3 years ago

Calculate the molar mass of cacl2

kirza4 [7]

Answer:

$\boxed {\boxed {\sf 110.98 \ g/mol}}$

Explanation:

The molar mass is the mass of a substance in grams per mole.

To find it, add the mass of each element in the compound. These masses can be found on the Periodic Table.

The compound given is:

$CaCl_2$

The compound has 1 Ca (calcium) and 2 Cl (chlorine).

Mass of Calcium

The molar mass of calcium is 40.08 g/mol
There is only one atom of Calcium in CaCl₂, so the number above is what we will use.

Mass of Chlorine

The molar mass of chlorine is 35.45 g/mol
There are two atoms of chlorine in CaCl₂, therefore we need to multiply the molar mass by 2.
35.45 * 2= 70.9 g/mol

Molar Mass of CaCl₂

Now, to find the molar mass, add the molar mass of 1 calcium and 2 chlorine.
40.08 g/mol + 70.9 g/mol =110.98 g/mol

The molar mass of CaCl₂ is <u>110.98 grams per mole. </u>

6 0

3 years ago

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

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3 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

<u>Answer:</u> The value of $K_{eq}$ is $4.84\times 10^{-5}$

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

<u>Initial:</u> 0.0280 0.0120

<u>At eqllm:</u> 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

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Leya [2.2K]

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