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4 years ago

Please help with this chemistry question above!

Chemistry

1 answer:

Lady_Fox [76]4 years ago

3 0

Hey there!

Your answer is B. CO₂.

This molecule consists of two polar bonds, but they cancel each other out because they are pointing away from each other.

Because the forces of the bonds cancel out, the molecule itself is nonpolar.

Hope this helps!

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using the equaation 2h2+o2-->2h2o if 10.0g of hydrogen are used in the presence of excess oxygen how many grams of water will

astra-53 [7]

Answer:

90g of H2O

Explanation:

2H2 + O2 —> 2H2O

First, we calculate the molar masses of H2 And H20.

Molar Mass of H2 = 2g/mol

Mass conc of H2 from the balanced equation = 2 x 2 = 4g

Molar Mass of H2O = 2 + 16 = 18g/mol

Mass conc of H2O from the balanced equation = 2x18 = 36g

From the equation,

4g of H2 produced 36g of H2O

Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O

7 0

3 years ago

Which substance is a diatomic molecule? <br> a. he <br> b. o2 <br> c. co2 <br> d. h2o?

BigorU [14]

b. o2
oxygen is diatomic because 1 molecule is made of 2 atoms of oxygen

4 0

3 years ago

Solve the 3 questions

Dmitrij [34]

Answer:

what kind of math is this

Explanation:

5 0

3 years ago

What is x if x - 2 =0

GaryK [48]

Answer:

Explanation:

8 0

3 years ago

Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20

KengaRu [80]

Answer:

$m_{Ag}=2,265.9g$

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

$\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]$

By knowing the moles of gold:

$n_{Au}=100g*\frac{1mol}{197g} =0.508mol$

It is possible to write the aforementioned formula in terms of the variable $x$ representing the moles of silver:

$20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]$

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

$x=n_{Ag}=21.0molAg$

So the mass is:

$m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g$

Best regards!

3 0

3 years ago