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3 years ago

Which statement describes one characteristic of an operating electrolytic cell?

Chemistry

1 answer:

kati45 [8]3 years ago

8 0

Answer:

Explanation:

The mechanism of this process: the electrolytic cell converts the electrical energy to chemical energy, using external energy source.

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A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6

Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of $H = 993 \frac{J}{kgC}$

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

$H = \frac{E}{M*(T2-T1)}$

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

$H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}$

Finally, the specific heat of this metal is

$H = 993 \frac{J}{kgC}$

The aluminium could be the metal, its specific heat is similar to that found in this problem.

Finally, we can conclude that this metal could be the aluminium with a specific heat of $H = 993 \frac{J}{kgC}$

7 0

3 years ago

Mnenie [13.5K]

Answer:

it should be the second one since London dispersion causes interaction between molecules.

hope I helped.

6 0

3 years ago

Carbon-14 has a half-life of 5,700 years.

Maksim231197 [3]

Answer:

$3.106\ \text{g}$

Explanation:

$t_{1/2}$ = Half-life of carbon = 5700 years

t = Time at which the remaining mass is to be found = 10400 years

$m_0$ = Initial mass of carbon = 11 g

Decay constant is given by

$\lambda=\dfrac{\ln2}{t_{1/2}}$

Amount of mass remaining is given by

$m=m_0e^{-\lambda t}\\\Rightarrow m=m_0e^{-\dfrac{\ln2}{t_{1/2}} t}\\\Rightarrow m=11e^{-\dfrac{\ln 2}{5700}\times 10400}\\\Rightarrow m=3.106\ \text{g}$

The amount of the substance that remains after 10400 years is $3.106\ \text{g}$ .

5 0

3 years ago

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$