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Alex17521 [72]
3 years ago
6

A sample of V weighs 61.9 grams. Will a sample of He that contains the same number of atoms weigh more or less than 61.9 grams

Chemistry
1 answer:
damaskus [11]3 years ago
6 0

Answer:

Weigh less than 61.9 g

Explanation:

The symbol V represents the element Vanadium and He represents Helium.

Generally, 1 mol of any atom contains 6.022 * 10^23 atoms. This means we have to find the number of moles of each elements.

For V;

Number of moles = Mass / Molar mass = 61.9 / 50.94 = 1.215 mol

1 mol = 6.022 * 10^23 atoms

1.215 = x

Number of atoms = 6.022 * 10^23 * 1.215 =7.317 * 10^23

For He;

Since equal number of moles contain equal number of atoms;

Number of mol of He = 1.215 mol

Number of moles = Mass / Molar mass

Mass = Molar mass * Number of moles = 4 * 1.215 = 4.86 g

This means it would weigh less.

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A compound has a percent composition of 15.24% sodium (molar mass = 22.99
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The empirical formula of the compound obtained from the question given is NaBrO₃

<h3>Data obtained from the question </h3>
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  • Bromine (Br) = 52.95%
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<h3>How to determine the empirical formula </h3>

The empirical formula of the compound can be obtained as illustrated below:

Divide by their molar mass

Na = 15.24 / 22.99 = 0.663

Br = 52.95 / 79.90 = 0.663

O = 31.81 / 16 = 1.988

Divide by the smallest

Na = 0.663 / 0.663 = 1

Br = 0.663 / 0.663 = 1

O = 1.988 / 0.663 = 3

Thus, the empirical formula of the compound is NaBrO₃

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Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride:
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In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.

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Which of the following is a pure substance
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How many moles of electrons must be transferred to plate out 110 g of manganese (MW ~ 55g/mol) from a solution of permanganate (
ololo11 [35]

Answer:

14 mol e⁻

Explanation:

Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese

8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)

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The molar mass of Mn is 55 g/mol.

110 g × 1 mol/55 g = 2 mol

Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn

According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.

2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻

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