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Keith_Richards [23]
2 years ago
14

The Reaction: _S+O2=_SO2 How many moles of sulfur must be burned to give 0.567 moles of SO2?

Chemistry
2 answers:
murzikaleks [220]2 years ago
8 0

Answer:

0.567mol SO_{2}

Explanation:

We have the reaction

S+ O_{2}\longrightarrow So_{2}

we balance the equation

S+ O_{2}\longrightarrow So_{2}

In this case the equation is already balanced

We use stoichiometric relations to solve it.

we know that 1 mol of sulfur produces 1 mol of SO_{2} (we know it by the coefficients of balancing of the equation, which in this are equal to 1)

Now how many moles of sulfur are needed to produce 0.567 moles of SO_{2}

1mol S\longrightarrow 1 mol SO_{2}\\ x\longrightarrow 0.567mol SO_{2}\\ x=\frac{0.567 mol SO_{2}.1molS}{1molSO_2} =0.567mol SO_{2}

Vesna [10]2 years ago
3 0
Moles ratio:

1 S<span> + 1 </span><span>O2</span><span> = 1 </span><span>SO<span>2
</span></span>
1 mole S -------------- 1 mole SO2
? moles S ------------ 0.567 moles SO2

0,567 x 1 / 1

= 0.567 moles of S


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A 5.0 g sample of calcium nitrate (Ca(NO3)2, M = 164) contaminated with silica (SiO2, M = 60.1) is found to contain 1.0 g calciu
zhenek [66]

Answer:

= 82%

Explanation:

Percentage purity is calculated by the formula;

% purity = (mass of pure chemical/total mass of sample) × 100

In this case;

1 mole of Ca(NO3)2 = 164 g

but; 164 g of Ca(NO3)2 = 40 g Ca

Therefore; mass of Ca(NO3)2 = 164 /40

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Thus;

% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100

                                    = (4.1 g/ 5 g) × 100

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4 0
2 years ago
Which of the following are benefits of Earth's atmosphere? Select all that apply Exchanges gases with the hydrosphere, recycling
VikaD [51]
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5 0
2 years ago
vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu
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Explanation:

Average atomic mass of the  vanadium = 50.9415 amu

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Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

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The atomic mass of isotope (II) of vanadium is 49.944 amu.

5 0
2 years ago
For some hypothetical metal, the equilibrium number of vacancies at 600°C is 1 × 1025 m-3. If the density and atomic weight of t
makvit [3.9K]

Answer:

\frac{N_{v}}{N}=1.92*10^{-4}

Explanation:

First of all we need to find the amount of atoms per volume (m³). We can do this using the density and the molar mass.

7.40 \frac{g}{cm^{3}}*\frac{1mol}{85.5 g}*\frac{6.023*10^{23}atoms}{1mol}*\frac{1000000 cm^{3}}{1m^{3}}=5.21*10^{28}\frac{atoms}{m^{3}}

Now, the fraction of vacancies is equal to the N(v)/N ratio.

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Therefore:  

The fraction of vacancies at 600 °C will be:

\frac{N_{v}}{N}=\frac{1*10^{25}}{5.21*10^{28}}  

\frac{N_{v}}{N}=1.92*10^{-4}

I hope it helps you!

 

7 0
2 years ago
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