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AlladinOne [14]
2 years ago
7

A 17.0 g sample of CaSO4, is found to contain 5.00 g of Ca and 7.99 g of O. Find the mass of sulfur in a sample of CaSO4 with a

Chemistry
1 answer:
TEA [102]2 years ago
8 0
For this question the answer is 32.065
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How many moles of water are formed when 22 mol of methane combusts?
icang [17]
Methane is a hydrocarbon which when burns in air (combustion) produces carbon dioxide and water. The equation for the reaction;
CH4 +2O2 = CO2 +2H2O
When one mole of methane combusts 2 moles of water are formed
Therefore; when 22 moles of methane combusts 44 moles of water are formed (22 ×2)
7 0
3 years ago
Carbon exists in nature as three isotopes. Given the atomic masses and relative abundance of the isotopes, what is the average a
Ivan

The most abundant carbon isotope is carbon-12.

The relative atomic mass of carbon is 12.011, which is extremely close to 12.0. This means that the masses C-13, and C-14 are practically negligible when contributing to the relative atomic mass of carbon.

the C-12 isotope makes up 98.9% of carbon atoms, C-13 makes up 1.1% of carbon atoms, and C-14 makes up just a trace of carbon atoms as they are found in nature.

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3 years ago
JUST ONE MULTIPLE CHOICE QUESTION!
Butoxors [25]
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Which method of heat transfer is due to the transfer of kinetic energy from one molecule to another?
pshichka [43]

Answer: Conduction

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3 0
3 years ago
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Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0
3 years ago
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