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kenny6666 [7]
3 years ago
15

Help please chemistry question

Chemistry
1 answer:
Gemiola [76]3 years ago
3 0

Answer:

D. K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}

Explanation:

The general form of an equilibrium constant expression is

K = \frac{[\text{Products}]}{[\text{Reactants}]}

In the equilibrium

HNO₂ ⇌ H⁺ + NO₂⁻

The products are H⁺ and NO₂⁻, and the reactant is HNO₂.

∴ K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}

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How many moles of ammonium ions are in 125 mL of 1.40 M NH4NO3 solution? ________ moles (give answer with correct sig figs in un
Sholpan [36]

The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:

Volume = 125 mL = 125 / 1000 = 0.125 L

Molarity = 1.40 M

<h3>Mole of NH₄NO₃ =? </h3>

Mole = Molarity x Volume

Mole of NH₄NO₃ = 1.40 × 0.125

<h3>Mole of NH₄NO₃ = 0.175 mole</h3>

Finally, we shall determine the number of mole of ammonium ion, NH₄⁺ in the solution. This can be obtained as follow:

NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)

From the balanced equation above,

1 mole of NH₄NO₃ contains 1 mole of NH₄⁺

Therefore,

0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺

Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

Learn more: brainly.com/question/25469095

3 0
2 years ago
What am I calculating for when solving for 17.0 mL of ethanol, density of ethanol= 0.94 g/mL
True [87]

Answer:

you are caculating for the mass/wight of ethanol

Explanation:

DxV=M

5 0
3 years ago
The following is a procedure that was theoretically performed by a student. Read through the procedure and answer the questions
yKpoI14uk [10]

Answer:

CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)

Explanation:

We have the reactants as calcium chloride and potassium carbonate. Recall that we are expecting that the reaction will yield a precipitate. We must keep that in mind as we seek to write its balanced chemical reaction equation.

So we now have;

CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)

Recall that the rule of balancing chemical reaction equation states that the number of atoms of each element on the right side of the reaction equation must be the same as the number of atoms of the same element on the left hand side of the reaction equation.

6 0
3 years ago
Help me this 2 . I’m not good at this sorry
emmainna [20.7K]

Left is respiratory

Right is circulatory

6 0
3 years ago
Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.
Anit [1.1K]

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molarity=\frac{n}{W_s}

where,

n = moles of solute

W_s = weight of solvent in kg

moles of acetone (solute) = 0.241

moles of water (solvent )= (1-0.241) = 0.759

mass of water (solvent )= moles\times {\text {Molar Mass}}=0.759\times 18=13.7g=0.0137kg

Now put all the given values in the formula of molality, we get

Molality=\frac{0.241}{0.0137kg}=17.6mole/kg

Therefore, the molality of solution is 17.6 mole/kg

3 0
3 years ago
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