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vazorg [7]
3 years ago
15

I really need help! If I have 17 moles of gas at a temperature of 67°C, and a pressure of 5.34 atmospheres, what is the volume o

f the gas? Enter the unit of your answer here. Enter it as a symbol, not words spelled out.
Chemistry
1 answer:
zalisa [80]3 years ago
5 0

Answer:

2

Explanation:

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H E L P <br> P L E A S E<br> ;-;
aleksandr82 [10.1K]

Answer:

because it decompose

Explanation:

yeast is an organic compound

7 0
3 years ago
Is aloe vera a living or non-living thing?
Afina-wow [57]

Answer:

It is a living plant

Explanation:

3 0
3 years ago
Select the true statements about hydrocarbons.
daser333 [38]

Answer:

a. Hydrocarbons have low boiling points compared to compounds of similar molar mass.

b. Hydrocarbons are hydrophobic.

d. Hydrocarbons are insoluble in water.

Explanation:

As we know that the hydrocarbons is a mix of carbon and hydrogen. In this the availability of the electronegative atom is not there that shows there is no bonding of the hydrogen plus it is dissolved. Also, the hydrocarbons is considered to be a non-polar but as compared to the water, water is a polar

In addition to this, the strong bond is no existed that shows the lower boiling points  

Therefore option A, B and D are right

8 0
3 years ago
Which metal will more easily lose an electron sodium or potassium?
sergey [27]
<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>
7 0
3 years ago
Calculate the DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. DH°f means delta or change
USPshnik [31]

Answer: +178.3 kJ

Explanation:

The chemical equation  follows:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]

We are given:

\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]

The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ

5 0
3 years ago
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