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3 years ago

What happens to the coefficients in a reaction when writing the equilibrium constant expression?

Chemistry

1 answer:

Taya2010 [7]3 years ago

6 0

The answer is D :))..

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Liquid X is known to have a lower viscosity and lower surface tension than Liquid Y Use these facts to predict the result of eac

Dima020 [189]

Liquid is safe i think hope this helps

8 0

2 years ago

By titration, it is found that 28.5 mL of 0.183 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the

maksim [4K]

Answer:

0.209M

Explanation:

M1V1=M2V2

(28.5 mL)(0.183M)=(25.0mL)(M)

M2= 0.209M

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

4 0

2 years ago

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

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

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

4 0

3 years ago

Why is it better to measure the percent change in mass<br> ratherthan just using the change in mass?

never [62]

Answer: The result is presented in proportion which gives a clearer understanding and accurate result.

Explanation: Percentage change in mass is the proportion of the initial mass of a substance changed after sometime. The results is presented as a percentage making it more accurate and can help to give future reference to weight calculations.

Change is Mass is the mass of a substance left after sometime mostly given in grams. It is not as accurate as percentage change in mass. It is generally better to show results in percentage change in mass as it gives a better understanding of what mass of a substance was lost after a given period or after application of energy like Heat or increased temperature.

6 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$