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natita [175]
3 years ago
10

Calculate the number of moles of potassium permanganate (KMnO4) corresponding to 230.8 g of the substance.

Chemistry
1 answer:
vlada-n [284]3 years ago
6 0

Answer:

The answer to your question is 1.46 moles of KMnO₄

Explanation:

Data

number of moles = ?

mass = 230.8 g

molecular mass of KMnO₄ = 39 + 55 + (16 x 4) = 158 g

Process

1.- Use proportions and cross multiplication to answer this problem.

                          158 g of KMnO₄ ----------------  1 mol

                           230.8 g of KMnO₄ -----------    x

                           x = (230.8 x 1) / 158

2.- Simplifying

                           x = 230.8 / 158

3.- Result

                          x = 1.46 moles

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"calculate the ratio of the velocity of hydrogen molecules to the velocity of carbon dioxide molecules at the same temperature"
Ne4ueva [31]

Answer: 1:4.69

Explanation:

The ratio can be expressed as:

Ua/Ub= √(Mb/Ma)

Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2

Mb is the molecular mass of CO2 = 44

Therefore

Ua/Ub= √(44/2)

Ua/Ub = 4.69

Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69

which implies hydogen is about 4.69 times faster than carbon dioxide.

8 0
3 years ago
A chemical equation does not give information about what
Lana71 [14]

A chemical equation does not give information about the following:

  • It usually does not give the "state of the substances". There are three states: Solid(s), liquid(q) and gas(vap).
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8 0
3 years ago
Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea
galina1969 [7]

Answer:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Oxidized: Cd

Reduced: Ag

Explanation:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Cd → Cd²⁺  +  2e⁻      Half reaction oxidation

1e⁻ + Ag⁺ → Ag           Half reaction reduction

Ag changed oxidation number from +1 to 0

Cd changed oxidation number from 0 to +2

Let's ballance the electrons

( Cd → Cd²⁺  +  2e⁻ ) .1

( 1e⁻ + Ag⁺ → Ag ) .2

Cd + 2e⁻ + 2Ag⁺  → 2Ag +  Cd²⁺  +  2e⁻

Finally the ballance equation is:

Cd(s) + 2AgNO₃(aq)  → Cd(NO₃)₂ (aq) + 2Ag(s)

4 0
3 years ago
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Dahasolnce [82]

Answer:

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Explanation:

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6 0
2 years ago
An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is
aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x (\frac{1 L}{1000 mL}) = .236 L

Pressure= 740 mm Hg x (\frac{1 atm}{760 mm Hg})= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= \frac{mass}{moles} or MM= \frac{m}{n}. They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n=\frac{PV}{RT}, where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n=\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}

n= 0.009 mol

Let's look back at MM= \frac{m}{n} and plug in what we know.

MM= \frac{0.443 g}{0.009 mol}

MM= 49.2 g/mol

3 0
2 years ago
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