All categories

3 years ago

What are the three kind of rocks that are apart of the rock cycle???

Chemistry

2 answers:

Afina-wow [57]3 years ago

8 0

Sedimentary, metamorphic, and igneous

Hope this helps! :)

pav-90 [236]3 years ago

5 0

Why, Hello There! The 3 Kinds of rocks that are apart of the rock cycle is...

1.Sedimentary

2.Metamorphic

And

3.Igneous

Hoped i helped!

You might be interested in

What effect does high temperature have on radish growth size?

Juliette [100K]

Radishes are a popular root vegetable known scientifically as Raphanus raphanistrum. Radishes come in a variety of colors, including black, red, purple, and white. They have been harvested for thousands of years.

3 0

3 years ago

Read 2 more answers

What is hydrogen ion concentration of an acid solution whose PH is 3.4?

meriva

It’s easy, if the PH of any acidic solution = -Log[H+], where [H+] is hydrogen ion concentration, multiply each term by (-1) then raise each term as a power to (10), so it will become like this:-
[H+] = 10^(-PH)

4 0

3 years ago

Someone help me Explain how non-genetic factors can affect phenotype. Give an example.

Yanka [14]

Answer:

one could be the weather.

phenotypes are physical characteristics of a person and there non genetic factors that play a role.

weather and environmental factors, your surroundings and orders related to the environment basically

6 0

3 years ago

List three properties of metals that nonmetals typically do not have

Law Incorporation [45]

Conductivity, malleability, and high melting points. Hope this helps :)

7 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

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:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

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 = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago