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WINSTONCH [101]
3 years ago
8

A 40 g sample of sucrose contains 16.8 g of carbon. what is the mass

Chemistry
1 answer:
NemiM [27]3 years ago
3 0

Answer:

Thus 100.00 g of sucrose contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen.

Explanation:

Hope this helps, leave a heart ❤️

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The diagram shows a model of the nitrogen cycle. which role do plants play in the nitrogen cycle?
kenny6666 [7]

Answer

D

Explanation:

They take up usable forms of nitrogen found in soil

3 0
3 years ago
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A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point consta
aalyn [17]

Answer:

- 0.99 °C ≅ - 1.0 °C.

Explanation:

  • We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.

<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>

4 0
3 years ago
The molar mass of copper(II) chloride (CuCl2) is 134.45 g/mol. How many formula units of CuCl2 are present in 17.6 g of CuCl2? 7
Sedaia [141]
7.88  10^22 formula units 
5 0
3 years ago
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What is 0.00000005 in scientific notation, rounded to two significant figures?
KIM [24]

Answer:

5×-10^8

That is the the scintific notation for 0.00000005

Answer=5×-10^8

7 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
2 years ago
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