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AVprozaik [17]
2 years ago
12

How many moles are in 2.5L of 1.75 M Na2CO3

Chemistry
2 answers:
ra1l [238]2 years ago
7 0

Answer:

4.4 mol.

Explanation:

Have a great day

Natasha_Volkova [10]2 years ago
4 0

Answer:

4.4 mol.

Explanation:

Hello!

In this case, since the formula for calculating the molarity is:

M=n/V

Whereas n stands for moles and V for the volume in liters; we can solve for n as shown below when we are given the volume and the molarity:

n=V*M

Thus, we plug in the given data to obtain:

n=2.5L*1.75mol/L=4.4mol

Best regards!

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In a solution of pure water, the dissociation of water can be expressed by the following: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) T
kakasveta [241]

Answer:

  • [H₃O⁺] = 2.90 × 10⁻¹⁰ M

Explanation:

1)<u><em> Ionization equilibrium equation: given</em></u>

  • H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>

  • Kw = 1.0 × 10⁻¹⁴

<u>3) Stoichiometric mole ratio:</u>

As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:

  • [H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M

  • ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷  × 1.0 × 10⁻⁷  = 1.0 × 10⁻¹⁴ M

<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>

Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:

  • Kw = 1.0 × 10⁻¹⁴ M², and

  • Kw = [H3O⁺] [OH⁻]

Then you can substitute the known values and solve for the unknown:

  • 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M

  • ⇒ [H₃O⁺]  = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M

As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.

6 0
3 years ago
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<span>Zn⁰ + 2H⁺ ------> Zn²⁺ + H2⁰

H⁺ ion has oxidation number +1.
Zn²⁺ ion has oxidation number +2.
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</span><span>Answer is
D. Zn and each hydrogen atom in H2</span><span>

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3 years ago
At room temperature (208C) and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha
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Mass of the nitrogen gas : 1.165g/L \times 0.560L = 0.6524g

Learn more about Mass and Density here:

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*PLEASE HELP ASAP*
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