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vampirchik [111]
1 year ago
11

What mass of H atoms is contained in 50.7 g of NH3? Answer in units of g.

Chemistry
1 answer:
Ratling [72]1 year ago
7 0

The mass of hydrogen present in 50.7 grams of ammonia is equal to 8.95 g.

<h3>What is a mole?</h3>

A mole is a scientific unit that is used to determine the huge number of quantities of atoms, molecules, ions, etc. The mass of the 1 mole of any element is called atomic mass and that of one mole of any compound is called molar mass.

The number of entities present in 1 mole was found to be equal to 6.023 × 10 ²³ per mole which is known as Avogadro’s constant.

Given, the mass of ammonia = 50.7 g

The mass of the one mole of ammonia = 17 g/mol

17 g of ammonia contain the mass of hydrogen = 3 g

50.7 g of ammonia contains a mass of hydrogen = (3/17) × 50.7 = 8.95 g

Learn more about the mole, here:

brainly.com/question/26416088

#SPJ1

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Because it goes through the process of sublimation.

Explanation:

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H2CO3(aq) + H200 H30 (aq) + HCO3 (aq).
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Answer:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

  • an equilibrium constant is, first of all, a fraction;
  • in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
  • in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
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  • only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.

Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

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3 years ago
A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula
Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

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1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

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8 0
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Answer:

My guess is b or c but its robably wrong

Explanation:

I just also need points sorry <3

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