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Gnom [1K]
3 years ago
6

What volume of oxygen gas at 20°C and 0.919 atm is

Chemistry
1 answer:
ehidna [41]3 years ago
5 0

Answer:

Volume of oxygen gas is 78.47L

Explanation:

Data;

Temperature (T) = 20°C = (20 + 273.15)K = 293.15K

Pressure (P) = 0.919atm

Mass (m) = 5g

Volume (v) = ?

Equation of reaction;

4Fe + 3O₂ → 2Fe₂O₃

From the equation of reaction, the number of moles of oxygen is 3.

To find the volume of oxygen produced, we'll have to use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = ideal gas constant = 0.082J/mol.K

T = temperature of the ideal gas

PV = nRT

V = nRT/ P

V = (3 × 0.082 × 293.15) / 0.919

V = 72.1149 / 0.919

V = 78.47L

Volume of the gas is 78.47L

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Aretes and horns

Explanation:

Aretes and horns are the landforms that are created where cirques meet. These landforms are products of glacier erosion in most temperate and polar regions.

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6 0
3 years ago
Where does the term carbohydrate come from? Explain by referring to the parts of this word
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A carbohydrate comes from a chain of carbon atoms with an H2O associated with each other 
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4 years ago
How many moles of calcium chloride are contained in a 333 gram sample?
Bogdan [553]

Answer & Explanation:

The molar mass of calcium chloride is 110.98 g/mol. We can use this information to solve this problem. We can set up our equation like this..

\frac{333 g(CaCl2)}{} *\frac{1mol(CaCl2)}{110.98g(CaCl2)}

Multiply straight across on the top and straight across on the bottom.

\frac{333}{110.98}

Now divide.

\frac{333}{110.98}=3.00

So, there are 3.00 moles of calcium chloride contained in a 33 gram sample which is answer choice D.

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3 years ago
Flocabulary circulation
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3 0
3 years ago
: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be
Jet001 [13]

Answer: New concentration of HC_{6}H_{5}O^{2-}_{7} is 0.23 M.

Explanation:

The given data is as follows.

     Moles of HC_{6}H_{5}O^{2-}_{7} = 0.06 mol

     Moles of H_{2}C_{6}H_{5}O_{7} = 0.08 mol

Therefore, moles of OH^{-} added are as follows.

    Moles of OH^{-} = 0.125 \times \frac{25}{1000}

                          = 0.003125 mol

Now, new moles of HC_{6}H_{5}O^{2-}_{7} = 0.06 + 0.003125

                     = 0.063125

Therefore, new concentration of HC_{6}H_{5}O^{2-}_{7} will be calculated as follows.

       Concentration = \frac{0.063125}{0.275}

                               = 0.23 M

Thus, we can conclude that new concentration of HC_{6}H_{5}O^{2-}_{7} is 0.23 M.

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