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Anna35 [415]
3 years ago
9

1000 cm3 of hydrogen gas (hydrogen molecules, H2) contains x number of molecules at room temperature and pressure. Determine the

number of atoms in 500 cm3 of radon gas (radon atoms) at the same temperature and pressure
Chemistry
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer: Number of atoms in 500cm^3 of radon gas (radon atoms) at the same temperature and pressure is \frac{x}{2}

Explanation:

According to avogadro's law, volume of a gas is directly proportional to the number of moles present when temperature and pressure is constant.:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

V_1 = Volume of the hydrogen gas = 1000cm^3

n_1 =  moles of hydrogen gas = \frac{xmolecules}{6.023\times 10^{23}molecules}

V_2=  Volume of the radon gas =500cm^3

n_1 =  moles of radon gas = ?

\frac{1000}{\frac{x}{6.023\times 10^{23}}}=\frac{500}{\frac{y}{6.023\times 10^{23}}}

y=\frac{x}{2}

Thus the number of atoms in 500cm^3 of radon gas (radon atoms) at the same temperature and pressure is \frac{x}{2}

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Example: Mg^{2+}+2e^-\rightarrow Mg

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Answer:

A. percentage mass of iron = 5.17%

percentage mass of sand  = 8.62%

percentage mass of water = 86.205%

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C. The step of separation of iron and sand

Explanation:

A. Percentage mass of the mixtures:

Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g

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B. Flow chart of separation procedure

(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand

C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.

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Answer:

The % yield of the reaction = 27.5 %

Explanation:

Step 1: Data given

Mass of Li = 12.7 grams

Mass of N2 = 34.7 grams

Actual yield of Li3N = 5.85 grams

Molar mass of  Lithium = 6.94 g/mol

Molar mass of N2 = 28 g/mol

Molar mass of LI3N = 34.83 g/mol

Step 2: The balanced equation:

6Li(s) + N2(g) → 2Li3N(s)

Step 3: Calculate moles of Lithium

Moles Li = mass Li / Molar mass Li

Moles Li = 12.7 grams / 6.94 g/mol

Moles Li = 1.83 moles

Step 4: Calculate moles of N2

Moles N2 = 34.7 g/ 28 g/mol

Moles N2 = 1.24 moles

Step 5: Limiting reactant

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

Lithium is the limiting reactant. It will completely be consumed (1.83 moles).

N2 is in excess. There will be consumed 1.83 / 6 = 0.305 moles

There will remain 1.24 - 0.305 = 0.935 moles

Step 6: Calculate moles of Li3N

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

For 1.83 moles Li, we'll have 1.83/3 = 0.61 moles of Li3N

Step 7: Calculate mass of Li3N

Mass Li3N =moles LI3N * Molar Mass LI3N

Mass Li3N = 0.610 moles * 34.83 g/mol

Mass Li3N = 21.2463 grams = Theoretical yield

Step 8: Calculate % yield

% yield = actual yield / theoretical yield

% yield = (5.85 / 21.2463)*100% = 27.5%

The % yield of the reaction = 27.5 %

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