Answer: A
Explanation: Mass and volume is the amount, amount doesn’t make the phase of a matter change.
Answer:

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]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
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](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
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Answer:
A. Fossils of organisms that lived in warm climates have been found in arctic environments.
Explanation:
One of the logic used by Geologists to justify plate movement over a region of space is that the fossils of organisms that lived in warm climates have been found in arctic environments.
- this is as a result of plate tectonics.
- organisms that are well adapted to warm climates in times past have their relics preserved as fossils.
- their fossils are no found today around the arctic environment which are cold
- this suggests that the plate must have drifted or moved
Fossil fuels and unnatural energy sources still dominate in the US, this influences many problems including global warming, climate change, and pollution. Burning fossil fuels is also a problem to human health. And unnatural resources won’t be there forever whereas, renewable sources could last a long time.
Answer:
This is site for English speakers. Этот сайт на английском, поэтому вопрос могут удалить
Explanation:
1. 2)
2. 3)
3. 4) Sr
4. 3)
5. 4)
6. 2)
7. 1)
8. 4)
9. 3)
10. 3)
11. SO3, H2SO4, Na2SO4
12.
A) оксид меди (II) 2) CuO
Б) хлорид меди(II) 4) CuCl2
В) сульфит меди (II) 3) CuSO 3
Г) гидроксид меди (II) 1) Cu(OH)2
13.
1. Fe+HCl= б) FeCl 2 +H 2
2.Fe+O2= в) Fe 3 O 4
3. Fe(OH) 3 = г)Fe 2 O 3 +H 2O
4. FeCl 2 +NaOH= а) Fe(OH) 2 +NaCl
14. 2Ca + O2 = 2CaO
CaO + H2O = Ca(OH)2
Ca(OH)2 + 2HCl = CaCl2 + 2H2O