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2 years ago

The chemical reaction in which ammonium sulfate and

Chemistry

1 answer:

Ganezh [65]2 years ago

5 0

Answer:

Answer and Explanation: In this reaction ammonium sulfate reacts with potassium hydroxide to give aqueous potassium sulfate, gaseous ammonia, and liquid water as the products. This is essentially an acid-base reaction where the ammonium ion is donating a hydrogen ion to the hydroxide ion from the potassium hydroxide.

Explanation:

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nignag [31]

ADSORPTION: The adhesion of a liquid or gas on the surface of a solid material, forming a thin film on the surface. Not to be confused with the process of absorption.

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3 years ago

I need HELP! What is a variable that meteorologist track to make weather forecast? The answers are either: air pressure or barom

larisa [96]

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3 years ago

Read 2 more answers

How many atoms are in 1.0 formula units of magnesium acetate?

Ray Of Light [21]

Answer:

Explanation:

Magnesium Acetate Mg(C2H3O2)2

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3 years ago

An electron has an uncertainty in its position of 471 pm .

olya-2409 [2.1K]

<u>Answer:</u>

$\Delta x . \Delta p \geq \frac{h}{4 \pi}$

∆x is the Uncertainty in the position of the particle

∆p is the Uncertainty in the momentum of the particle

$\Delta p=\frac{h}{4 \pi \times m \times \Delta x}$

$=\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} s^{-1}}{4 \times 3.14 \times 9.109 \times 10^{-31} \mathrm{kg} \times 471 \times 10^{-12} \mathrm{m}}$

$=\frac{6.626 \times 10^{-34} \mathrm{ms}^{-1}}{53887 \times 10^{-43}}$

$=1.23\times10^5 ms^{-1}$ is the answer

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3 years ago

If hydrochloric acid is obtained commercially at a concentration of 12.1M, how many milliliters of 12.1M HCl(aq) must be used to

rewona [7]

Answer:

$V_1=82.6mL$

Explanation:

Hello there!

In this case, according to this question, we will need to deal with this dilution problem, because it is asking for the volume of a 12.1-M stock solution of HCl. In such a way, we can use the following equation, under the assumption of no change in the number of moles in the solution:

$M_2V_2=M_1V_1$

Thus, we solve for the initial volume, V1, as shown below:

$V_1=\frac{M_2V_2}{M_1}$

And plug in the initial concentration and final concentration and volume to obtain:

$V_1=\frac{2000mL*0.500M}{12.1M}\\\\V_1=82.6mL$

Regards!

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3 years ago