What property of a solution stays the same after its diluted

What pressure will be exerted by 0.675 moles of a gas at 25*C if it is in a 0.750-L container?

Westkost [7]

PV=nRT

P=nRT/V

P=[(0.650mol)(0.08206)(298K)]/(0.750L)=21.2atm

6 0

3 years ago

Select the correct structure that

Leto [7]

Answer:

c) both

Explanation:

hey, could you check the question once because I suppose it's 2-hexyne when there's a triple bond and 2-hexene when there's a double bond between Carbon atoms. As of the question asked, it's c) both but it's 2-hexene due to presence of double bond between Carbon atoms.

do tell me as well. have a good day :-))

6 0

3 years ago

The AE of a system that releases 12.4 J of heat and does 4.2 J of work on its surroundings It is_______ jA. 16.6 B. 12.4 C. 4.2

andre [41]

ANSWER

EXPLANATION

Given that

The energy released by the system is 12.4J

Work done on the surrounding is 4.2J

Follow the steps below to find the change in energy

In the given data, energy is said to be released to the surroundings

Recall, that exothermic reaction is a type of reaction in which heat is released to the surroundings. Hence, change in enthalpy is negative

Step 1; Write the formula for calculating change in energy

$\Delta E\text{ }=\text{ q }+\text{ w}$

Since heat is released to the surrounding, then q = -12J

Recall, that work done by the system on the surroundings is always negative

Hence, w = -4.2J

Step 2; Substitute the given data into the formula in step 1

$\begin{gathered} \text{ }\Delta E\text{ = q + w} \\ \text{ }\Delta E\text{ }=\text{ -12.4 }+\text{ \lparen-4.2\rparen} \\ \text{ }\Delta E\text{ = -12.4 - 4.2} \\ \text{ }\Delta E\text{ }=\text{ -16.6J} \end{gathered}$

Therefore, the change i

3 0

1 year ago

Pubhvcghbb<br> Jgyukbcu?

tino4ka555 [31]

Answer:

WHAT??

Explanation:

3 0

3 years ago

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Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i

g100num [7]

It's simple, just follow my steps.

1º - in 1 L we have $0.0100~g$ of $BaCO_3$

2º - let's find the number of moles.

$\eta=\frac{m}{MM}$

$\eta=\frac{0.0100}{197.3}$

$\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}$

3º - The concentration will be

$C=5.07\times10^{-5}~mol/L$

But we have this reaction

$BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}$

This concentration will be the concentration of $Ba^{2+}~~and~~CO_3^{2-}$

$K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}$

considering $[BaCO_3]=1~mol/L$

$K_{sp}=[Ba^{2+}][CO_3^{2-}]$

and

$[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L$

We can replace it

$K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})$

$K_{sp}\approx25.70\times10^{-10}$

Therefore the $K_{sp}$ is:

$\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}$

4 0

4 years ago

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