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3 years ago

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HELP!!!!!

2 answers:

faltersainse [42]3 years ago

5 0

Answer:

5.17.

Explanation:

<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴. </em>

[OH⁻] = 1.5 x 10⁻⁹ M.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(1.5 x 10⁻⁹ M) = 6.66 × 10⁻⁶ M.

∵ pH = - log[H₃O⁺]

<em>∴ pH = - log(6.66 × 10⁻⁶ M) = 5.17. </em>

Elena-2011 [213]3 years ago

4 0

Answer:

5.18

Explanation:

I hope this helps ;)

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Answer: The concentration of $Cl^-$ ions in the resulting solution is 1.16 M.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

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$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $AlCl_3$

We are given:

$n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL$

Putting all the values in above equation, we get

$M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M$

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3 years ago

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3 years ago

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$\qquad\qquad\huge\underline{{\sf Answer}}$

Here we go ~

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2 years ago

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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:

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

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

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

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

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

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

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

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

Final temperature is cleared from this expression:

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

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

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago