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3 years ago

What was formed when you dropped the rock to the sand?

Chemistry

2 answers:

castortr0y [4]3 years ago

6 0

Answer: 1. rock stone gase from developing the other difference

Rufina [12.5K]3 years ago

3 0

Answer:

rock stone gase from developing the other difference.

Explanation:

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Vishwanath had some money. he spent 3 upon 4 part of money to buy goods for his birthday,1 upon 5 part of money give to his sist

lord [1]

Answer:

The correct answer is - 800.

Explanation:

Given:

Total amount = ? or assume x

spend in buying birthday item = 3/4 of x

given to sister = 1/5 of x

remaining to mother = 40

solution:

the remaning amount = x- (3x/4+x/5) = 4=

=> x- 19x/20 = 40

=> x = 20*40

=> x = 800

thus, the correct answer is = 800

8 0

3 years ago

If the molar mass of the compound in problem 1 is 110 grams/mole, what is the molecular formula? With work?

tatyana61 [14]

C6H6O2 is the answer

3 0

3 years ago

Identify which sets of Quantum Numbers are valid for an electron. Each set is ordered (n,ℓ,mℓ,ms).

skad [1K]

3,2,0,1/2

1,0,0,1/2

3,2,2,1/2

4,2,1,1/2

2,1,-1,-1/2

THEY ARE VALID

3 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]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 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:

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

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

How many centimeters are in 5.6 x 10-2 meters?

drek231 [11]

Answer:

0.5

Explanation:

8 0

3 years ago