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Dmitry [639]
3 years ago
6

When did Darwin publicly share his views of evolution?

Chemistry
1 answer:
Veronika [31]3 years ago
4 0

Answer:

It was not until he was 50 years old, in 1859.

Explanation:

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If liquid water is exposed to normal atmospheric pressure, what needs to change in order to change its state of matter?
inn [45]

Answer: Option (b) is the correct answer.

Explanation:

State of a substance changes when heat is provided to a substance.

This is because when we heat water then intermoleclar forces present within its molecules tend to break down. Due to this molecules start to move away from each other.

As a result, kinetic energy of molecules increases and they collide rapidly. Hence, solid state of water changes into liquid state and upon excessive heating liquid state of water changes into vapor state.

Thus, we conclude that temperature of water needs to change in order to change its state of matter.

6 0
3 years ago
Read 2 more answers
Be sure to answer all parts. Write a balanced equation (including physical states) for the following reaction: Sodium carbonate,
aalyn [17]

Answer:

I think it would be:

NaCO3 (s)-->Na2O (s) + CO2 (g)

3 0
3 years ago
Ca3(PO4)2 + 3H2SO4  3CaSO4 + 2H3PO4.
Dominik [7]
If anything isn’t obvious, don’t hesitate to ask me
Check the attached photo out...

4 0
2 years ago
An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid,
natima [27]

Answer:

The answer is "Option B".

Explanation:

\to CH_3COOH + NaOH \longleftrightarrow  CH_3COONa + H_2O\\\\\to CH_3COONa + NaOH\longleftrightarrow CH3COONa\\\\\therefore \ mol\  NaOH = (5 \ E-3\  L)\times(0.10 \ \frac{mol}{L}) = 5 \ E-4\ mol\\\\

\to mol\ CH_3COOH = (0.05 \ L)\times(0.20 \frac{mol}{L}) = 0.01 \ mol\\\\\to C \ CH_3COOH = \frac{(0.01 \ mol - 5 \ E-4\ mol) }{(0.105 \ L)}\\\\\to C \ CH_3COOH = 0.0905 \ M\\\\\therefore \ mol \ CH_3COONa = (0.05\  L )\times (0.20 \ \frac{mol}{L}) = 0.01 \ mol\\\\

\to C \ CH_3COONa =  \frac{(0.01\  mol + 5 \ E-4\  mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\

\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\  E-6 - 1.75\  E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to  [H_3O^+] = 1.5835\  E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to  pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7

5 0
3 years ago
Which best represents a heterogeneous mixture?
iragen [17]
The best example of a heterogenous mixture is sand at the beach. This is a good example as the sand and water do not chemically come together to create a new mixture and the composition of both elements stay constant
3 0
3 years ago
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