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melisa1 [442]
3 years ago
11

What is the mass of sodium (Na) in 50 grams of table salt (NaCl)? Show your work.

Chemistry
2 answers:
Margaret [11]3 years ago
8 0

Answer:

19 g

Explanation:

Data Given:

Sodium Chloride (table salt) = 50 g

Amount of sodium (Na) = ?

Solution:

Molecular weight calculation:

NaCl = 23 + 35.5

NaCl = 58.5 g/mol

Mass contributed by Sodium = 23 g

calculate the mole percent composition of sodium (Na) in sodium Chloride.

Since the percentage of compound is 100

So,

Percent of sodium (Na) = 23 / 58.5 x 100

Percent of sodium (Na) = 39.3 %

It means that for ever gram of sodium chloride there is 0.393 g of Na is present.

So,

for the 50 grams of table salt (NaCl) the mass of Na will be

mass of sodium (Na) = 0.393 x 50 g

mass of sodium (Na) = 19 g

Arturiano [62]3 years ago
3 0

Answer:

1.20 on Plato

Explanation:

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What is the predominant intermolecular force in the liquid state of each of these compounds: hydrogen fluoride (HF), carbon tetr
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Answer:

HF - hydrogen bonding

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NF3 - Dipole-dipole

Explanation:

Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom such as fluorine, chlorine nitrogen, oxygen etc. Hence the dominant intermolecular force in HF is hydrogen bonding.

CBr4 is nonpolar because the molecule is tetrahedral and the individual C-Br dipole moments cancel out leaving the molecule with a zero dipole moment hence the dominant intermolecular force are the dispersion forces.

NF3 has a resultant dipole moment hence the molecules are held together by dipole-dipole interaction.

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2 years ago
Calculate the molarity of hydroxide ion in an aqueous solution that has a poh of 5.00
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Answer:

1.00 x 10^-5 M

Explanation:

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4 0
3 years ago
the chloride of a metal M contains 47.25% of metal 1.0 gram of metal would be displaced from a compound by 0.88 gram of another
GenaCL600 [577]

Answer:

The equivalent weight of M is approximately 31.8 g

The equivalent weight of N is approximately 27.98 g

Explanation:

The given parameters are;

The percentage of the the metal M in in the chloride = 47.25%

Where by the chemical formula for the metal chloride is MClₓ, we have;

47.25% of the mass of MClₓ = Mass of M = W

Therefore, we have;

\dfrac{0.4725}{W} = \dfrac{1}{W + 35.5 \cdot x}

0.4725 × (W +  35.5·x) = W

0.4725·W + 0.4725×35.5×x = W

W - 0.4725·W  = 16.77·x

0.5275·W = 16.77·x

W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M

The equivalent weight of M = 31.799 ≈ 31.8 g

Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g

The equivalent weight of N = 27.98 g.

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Ostrovityanka [42]

Explanation: This is a reaction of oxidation of H_2O_2 in the presence of acidified KMnO_4. Acidified KMnO_4 is a strong oxidizing agent.

To balance out the H^+ on the reactant side, we write H_2O on the product side.

Balancing out the following reaction gives us:

2MnO_4^-+6H^++5H_2O_2\rightarrow 2Mn^{2+}+8H_2O+5O_2

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