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2 years ago

Tijuan mixed a solid and a liquid together in a beaker

1 answer:

Nutka1998 [239]2 years ago

5 0

Tijuan mixed a solid and a liquid together in a beaker. He determined that a chemical change occurred. ... After a week, he observed that only solid salt remained in the cup and the mass had decreased I hope this helped! In some way

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Which is the product of that reaction

ki77a [65]

Answer:

Explanation:

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2 years ago

Isaac Newton’s investigations of gravity explained which truth?

umka2103 [35]

<span>C. Gravity acts on all objects in the universe.
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2 years ago

Read 2 more answers

Identify the products formed in this Brønsted-Lowry reaction.<br><br> HPO2−4+NO−2↽−−⇀acid+base

klemol [59]

Answer: The products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Explanation:

According to Bronsted-Lowry, acids are the species which donate hydrogen ions to another specie in a chemical reaction.

Bases are the species which accept a hydrogen ion upon chemical reaction.

For example, $HPO^{2-}_{4} + NO^{-}_{2} \rightleftharpoons HNO_{2} + PO^{2-}_{4}$

Here, the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Thus, we can conclude that the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

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3 years ago

What is the mass of 0.46 mol of MgCl2

Ronch [10]

Answer:

43.7

Explanation:

mole is equal mass concentration/ moler mass of MgCl²

8 0

3 years ago

A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th

Kisachek [45]

Answer:

$pH=4.1$

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

$pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1$

Regards.

3 0

3 years ago