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3 years ago

A catalyst will

Chemistry

1 answer:

scZoUnD [109]3 years ago

7 0

Answer:

increase the chemical rate

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Approximately, how many national parks are there in the United States? a. 25 b. 50 c. 120 d. 474

Len [333]

B, I would believe. There are 58, but B is the closest we can get.

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3 years ago

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PLS HURRYThe moon does not light up by itself. Instead, it reflects light from the sun. Question 10 options: True False

alukav5142 [94]

Answer:

awnser is true took the test

Explanation:

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3 years ago

How long would it take to produce enough aluminum to make a case (24 cans) of aluminum soft drink cans if each can used 3 g of a

kozerog [31]

Answer : It takes time to produce 3 g of aluminium is, 165 seconds.

Explanation :

As we are given that the mass of aluminium is, 3 grams. Now we have to calculate the total mass of aluminum.

Total mass of aluminum = 3g × 24 = 72 g

Now we have to calculate the moles of aluminum.

$\text{Moles of Al}=\frac{\text{Mass of Al}}{\text{Molar mass of Al}}$

Molar mass of Al = 27 g/mol

$\text{Moles of Al}=\frac{72g}{27g/mol}=2.67mol$

As, 1 mole of Al has 3 Faradays

So, 2.67 mole of Al has = $2.67\times 3=80.1$ Faradays

and,

Charge = $80.1\text{ Faradays}\times \frac{96500\text{ coulombs}}{1\text{ Faradays}}=7.73\times 10^6\text{ coulombs}$

Current = $5.00\times 10^4A\times \frac{93.9}{100}=46950A$

Now we have to calculate the time.

$Time=\frac{Charge}{Current}=\frac{7.73\times 10^6}{46950}=164.6s\aaprox 165s$

Hence, it takes time to produce 3 g of aluminium is, 165 seconds.

5 0

3 years ago

The freezing temperatures for water for Celsius and Fahrenheit scales are 0ºC and 32ºF. The boiling temperatures for water are 1

ddd [48]

Answer:

Therefore the required function is

$C= \frac{5}{9} (F-32)$

Therefore 25°C=57°F

Explanation:

F denotes temperature of Fahrenheit and C denotes temperature of Celsius.

$\frac{C-0}{100-0} =\frac{F-32}{212-32}$

$\Rightarrow C= \frac{100}{180} (F-32)$

$\Rightarrow C= \frac{5}{9} (F-32)$

Therefore the required function is

$C= \frac{5}{9} (F-32)$

Putting C=25°C in above equation

$25=\frac{5}{9} (F-32)$

$\Rightarrow 25 \times \frac{9}{5} = F-32$

⇒45 =F-32

⇒F=32+45

⇒F=57

Therefore 25°C=57°F

7 0

3 years ago

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC

Salsk061 [2.6K]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm$

Putting values in above equation, we get:

$K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.6 kJ/mol

7 0

3 years ago