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spin [16.1K]
3 years ago
7

a sample of CO2 occupies a volume of 280ml at a pressure of 1.3 atm and a temperature of 18 degrees celsius, what volume will th

e gas occupy at a temp of 35 degrees celsius and a pressure of 3.0 atm
Chemistry
1 answer:
monitta3 years ago
7 0

We can calculate the new volume of the gas using the Combined Gas Law:

(P1 x V1) / T1 = (P2 x V2) / T2

The initial volume, pressure, and temperature were 280 mL, 1.3 atm, and 291.15 K (changing the temperature into Kelvin is necessary), and the final volume, pressure, and temperature is V2, 3.0 atm, and 308.15 K. Plugging these values in and solving, we find that:

(P1 x V1) / T1 = (P2 x V2) / T2

(1.3 atm x 280 mL) / 291.15 K = (3.0 atm x V2) / 308.15 K

V2 = 128.42 mL

This makes sense considering the conditions, a small increase in temperature would make the gas expand but a significant increase in the pressure would cause the volume to decrease.

Hope this helps!

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5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points
labwork [276]

Answer: (A) H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B) H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of H_{3}O^{+} in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

       Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)

 NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)

Hence, net ionic equation will be as follows.

       H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B)  When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

     NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)

The net ionic equation is as follows.

        H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

7 0
3 years ago
At catalyst is ______.
AlexFokin [52]
Heya!!!

Answer to your question:

A catalyst is ______
B. Not used up in a reaction.

Catalyst do not change the amount of reactants or products or itself get used but it just change the rate of reaction.

Hope it helps *_*
4 0
4 years ago
Name the compound CH3-CH3
Mkey [24]

Answer:

Ethane

Explanation:

CH3-CH3 is an alkane by the name of ethane. In organic chemistry there is a systematic way of naming compounds according to the prefix (eth) and the suffix (ane.)

The prefixes vary according to the number of carbon atoms:

1 C = meth

2 C = eth

3 C = prop

4 C = but

3 0
3 years ago
An egg is fried is that a physical or a chemical change
BigorU [14]
<span>Chemical change.........................................................................</span>
3 0
3 years ago
Read 2 more answers
The reaction between carbon tetrachloride, CCl4, and water, H2O, to form carbon dioxide, CO2, and hydrogen chloride, HCl, has a
xxTIMURxx [149]

Answer:

High activation energy is the reason behind unsuccessful reaction.

Explanation:

There are two types of reaction: (1) thermodynamically controlled reaction and (2) kinetically controlled reaction.

Thermodynamically controlled reaction are associated with change in enthalpy during reaction. More negative the enthalpy change, more favored will be the reaction.

Kinetically controlled reaction are associated with activation energy of a reaction. The lower the activation energy value, the more rapid will be the reaction.

Here, reaction between CCl_{4} and H_{2}O is thermodynamically favored due to negative enthalpy change but the high activation energy does not allow the reaction to take place by simple mixing.

6 0
3 years ago
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