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kumpel [21]
3 years ago
15

A student was given a pale blue transparent liquid in a beaker. The student gently heated the solution until the liquid had vapo

rized. Some blue crystals remained on the bottom of the beaker, why.
Chemistry
1 answer:
Vikki [24]3 years ago
3 0
This is so as the liquid you are heating is copper (II) sulfate (I think so) so when you heat it to saturation, there will still be some water molecules left behind, which will allow copper (II) sulfate crystals to be formed since there is water of crystallisation. so the formula is (CuSO4.5H2O).

Hence, if you heat it for a longer period of time when all the water has evaporated, you will obtain a white powder (CuSO4) as crystals cannot form without water of crystallisation
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What is the charge on an atom after it gains two electrons during the formation of a bond?
scoundrel [369]

Answer:

negative or neutral

Explanation:

electrons have a negative charge

  • the atom can be balanced by the negative charge if it was at two protons before

6 0
3 years ago
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Some one help me please I'm stuck Show using two conversion factors how you would convert from 0.020 kg into mg.
julia-pushkina [17]

Answer:

=2.0x10^4mg

Explanation:

Hello there!

In this case, when performing units conversions involving two proportional factors we need to make sure we first convert to the base unit and then to the target one; thus, since 1 kg = 1000 g and 1 g = 1000 mg, we set up the following expression:

=0.020kg*\frac{1000g}{1kg} *\frac{1000mg}{1g} \\\\=2.0x10^4mg

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4 0
3 years ago
Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1
givi [52]

According to Hasselbach-Henderson equation:

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

Here, [A^{-}] is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is CH_{3}COOK and acid is CH_{3}COOH thus, Hasselbach-Henderson equation will be as follows:

pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{2}{1}=5.06

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{3}=4.28

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{5}{1}=5.45

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{1}=4.76

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{10}=3.76

Therefore, pH of solution is 3.76.

4 0
3 years ago
What decides the pitch of sound
dmitriy555 [2]
The vibration objects
3 0
3 years ago
In the reaction 2H2(g) + O2(g) ® 2H2O(g), what is the volume ratio of H2 to H2O?
densk [106]
By looking at the stochiometry, we can conclude that the ratio is 1 :1

2 moles of H2 react to form 2 moles of H2O

hope this helps
3 0
3 years ago
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