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3 years ago

HELP!!! PLEASE!!!

1 answer:

8 0

Molarity is expressed as the number of moles of solute per volume of the solution. We solve the problem above as follows:

0.1000 mol Mg(NO3)2 / L (.1 L) ( 148.3 g / mol ) = 1.483 g Mg(NO3)2

0.1000 mol Sr(NO3)2 / L (.1 L) ( 211.63 g / mol ) = 2.116 g Sr(NO3)2

Hope this answers the question. Have a nice day.

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If solid NaCl is added to a saturated water solution of PbCl2 at 20o C, a precipitate is formed. How would this affect the value

Vadim26 [7]

For the question above, the answer would be C. The Ksp remains the same. The common ion effect will change the concentrations of the ions in thesolution, but the value of Ksp is CONSTANT at a constant temperature thus Ksp does not change.

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4 years ago

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How does activation energy affect a chemical reaction ?

PilotLPTM [1.2K]

Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) increases. Figure 3: Lowering the Activation Energy of a Reaction by a Catalyst.

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3 years ago

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Lavoisier made an important contribution to chemistry by _____.

Arada [10]

My answer would be D, 100% sure.

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3 years ago

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As the amount of charge on two objects increases, the strength of the electrical force between the objects

Aleks [24]

Answer:

Increase is the answer

Explanation:

Increase is the answer hopes this helps you

6 0

3 years ago

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A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L

Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0

3 years ago