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daser333 [38]
2 years ago
9

Which compound reacts with an acid to produce water and a salt

Chemistry
2 answers:
finlep [7]2 years ago
4 0
KOH is the only base, and any neutralization reaction produces a salt and water
Hatshy [7]2 years ago
3 0
I'm not exactly sure which one but I do know that an acid and a base react in a aqueous solution to form water, so i would probably eliminate the ones that aren't aqueous solutions.
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When temperatures are below freezing, the temperature at which air becomes saturated leading to the formation of frost is the
12345 [234]

Answer:

The answer is Frost Point.

Explanation:

The temperature to which the air must be cooled, with constant pressure, to reach saturation (in relation to liquid water), is called the dew point. The dew point gives a measure of the water vapor content in the air. The higher, the greater the concentration of water vapor in the air. However, when cooling produces saturation at a temperature of 0 ° C or less, the temperature is called a frost point. The water vapor is deposited as frost on a surface whose temperature is below the dew point.

4 0
3 years ago
Calculate the [oh−] in a solution with a ph of 12.52.
BaLLatris [955]
PH + pOH = 14

12.52 + pOH = 14

pOH = 14 - 12.52

pOH = 1.48

[OH⁻] = 10^ -pOH

[OH⁻] = 10 ^- 1.48

[OH⁻] = 0.033 M
6 0
3 years ago
Who proposed the "Law of Octaves" for organizing events?
djyliett [7]

b

Explanation:

February 7, 1863, was the day John Newlands published a paper outlining what would be known as “The Law of Octaves”. Newlands discovered if he ordered the known elements by increasing atomic weight, the chemical properties of the elements would be similar for every eighth group

5 0
3 years ago
Give two examples of steroids
Sonbull [250]
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3 0
2 years ago
An aqueous solution contains 0.29 M of benzoic acid (HA) and 0.16 M of sodium benzoate (A-). If the pH of this solution was meas
Inga [223]

Answer:

pKa = 4.89.

Explanation:

We can solve this problem by using the <em>Henderson-Hasselbach equation</em>, which states:

pH = pKa + log \frac{[A^-]}{[HA]}

In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

We <u>input the given data</u>:

4.63 = pKa + log \frac{0.16}{0.29}

And <u>solve for pKa</u>:

pKa = 4.89

3 0
3 years ago
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