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Oksi-84 [34.3K]
3 years ago
10

A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?

Chemistry
1 answer:
prohojiy [21]3 years ago
3 0

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

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Answer:

2,3,6,1

2,3,6,1

Explanation:

The unbalanced reaction expression is given as:

  AlBr₃    +    K₂SO₄  →    KBr  + Al₂(SO₄)₃

We need to balanced this reaction equation. Our approach is a mathematical method where we assign variable a,b,c and d as the coefficients.

 aAlBr₃    +    bK₂SO₄  →    cKBr  + dAl₂(SO₄)₃

Conserving Al;  a  = 2d

                   Br:  3a  = c

                   K:   2b = c

                   S:   b  = 3d

                   O:  4b  = 12d

Let a  = 1, c = 3, d  = \frac{1}{2}  b  = \frac{3}{2}  

 Multiply through by 2 to give;

     a  = 2, b = 3, c = 6 and d  = 1

 2AlBr₃    +    3K₂SO₄  →    6KBr  + Al₂(SO₄)₃

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The geometry on</span><span> how two acetonitrile molecules would interact with each other is as follows:

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Digiron [165]

The number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

<h3>HOW TO CALCULATE NUMBER OF ATOMS?</h3>

The number of atoms in a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

The number of moles in the gold (Au) can be calculated by dividing the mass of gold by its molar mass (196.97g/mol).

no. of moles = 23.5g ÷ 196.97g/mol

no. of moles = 0.119mol

Number of atoms in Au = 0.119 × 6.02 × 10²³

no. of atoms = 7.18 × 10²² atoms.

Therefore, the number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

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3 0
2 years ago
In an acid-base neutralization reaction 38.74 ml of 0.500 m potassium hydroxide reacts with 50.00 ml of sulfuric acid solution.
DaniilM [7]

Answer:

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Explanation:

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The no. of millimoles of acid is equal to that of the base at the neutralization.

<em>∴ (XMV) KOH = (XMV) H₂SO₄.</em>

X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),

M is the molarity.

V is the volume.

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X = 1, M = 0.5 M, V = 38.74 mL.

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X = 2, M = ??? M, V = 50.0 mL.

∴ M of H₂SO₄ = (XMV) KOH/(XV) H₂SO₄ = (1)(0.5 M)(38.74 mL)/(2)(50.0 mL) = 0.1937 M ≅ 0.2 M.

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