When an aqueous solution of magnesium nitrate is mixed with an aqueous solution of potassium carbonat?

PLEASE HELP QUICK I NEED 14-16 (JUST 6 PROBLEMS) ANSWERED AND FULL WORK SHOWN FOR FACTOR LABEL METHOD ITS 100 POINTS AND I WILL

Salsk061 [2.6K]

Answer:

14. 21.51 liters

15. 12.9 centimeters

16. 3400 milliliters

Explanation:

I don't really know how to show the work

5 0

3 years ago

The standard free energy of activation of one reaction A is 95.00 kJ mol–1 (22.71 kcal mol–1). The standard free energy of activ

diamong [38]

Answer:

The answer to the questions are as follows

Reaction B is 4426.28 times faster than reaction A

(b) Reaction B is faster.

Explanation:

To solve the question we are meant to compare both reactions to see which one is faster

The values of the given activation energies are as follows

For A

Ea = 95.00 kJ mol–1 (22.71 kcal mol–1) and

for B

Ea = 74.20 kJ mol–1 (17.73 kcal mol–1)

T is the same for both reactions and is equal to 298 k

Concentration of both reaction = 1M

The Arrhenius Law is given by

k = $Ae^{\frac{-E_{a} }{RT} }$

Where

k = rate constant

Ea = activation energy

R = universal gas constant

T = temperature (Kelvin )

A = Arrhenius factor

Therefore

For reaction A, the rate constant k₁ is given by k₁ = $Ae^{\frac{-95000}{(8.314)(298)} }$

And for B the rate constant k₂ is given by k₂ = $Ae^{\frac{-74200 }{(8.314)(298)} }$

k₁ = A×2.225×10⁻¹⁷

k₂ = A×9.850×10⁻¹⁴

As seen from the above Reaction B is faster than reaction A by (A×9.850×10⁻¹⁴)/(A×2.225×10⁻¹⁷) or 4426.28 times

3 0

3 years ago

A chemist fills a reaction vessel with 3.82 atm methanol (CH,OH) gas, 7.56 am oxygen (O2) gas, 5.29 atm carbon dioxide (CO2) gas

alisha [4.7K]

<u>Answer:</u> The Gibbs free energy of the given reaction is $1.379\times 10^3kJ$

<u>Explanation:</u>

The equation used to calculate Gibbs free energy change is of a reaction is:

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

For the given chemical reaction:

$2CH_3OH(g)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(CO_2(g))})+(4\times \Delta G^o_f_{(H_2O(g))})]-[(2\times \Delta G^o_f_{(CH_3OH(g))})+(3\times \Delta G^o_f_{(O_2(g))})]$

We are given:

$\Delta G^o_f_{(H_2O(g))}=-228.57kJ/mol\\\Delta G^o_f_{(CO_2(g))}=-394.36kJ/mol\\\Delta G^o_f_{(CH_3OH(g))}=-161.96kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-394.36))+(4\times (-228.57))]-[(2\times (-161.96))+(3\times (0))]\\\\\Delta G^o_{rxn}=-1379.08kJ/mol$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln Q_{p}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = -1379.08 kJ/mol = -1379080 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/K mol

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

$Q_{p}$ = Ratio of concentration of products and reactants = $\frac{(p_{CO_2})^2(p_{H_2O})^4}{(p_{CH_3OH})^2(p_{O_2})^3}$

$p_{CO_2}=5.29atm\\p_{H_2O}=3.89atm\\p_{CH_3OH}=3.82atm\\p_{O_2}=7.56atm$

Putting values in above expression, we get:

$\Delta G=-1379080J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(5.29)^2\times (3.89)^4}{(3.82)^2\times (7.56)^3}))\\\\\Delta G=-1379039J=1379.039kJ=1.379\times 10^3kJ$

Hence, the Gibbs free energy of the given reaction is $1.379\times 10^3kJ$

5 0

3 years ago

What cannot be broken down into other substances?

MatroZZZ [7]

Answer:

element

Explanation:

8 0

4 years ago

The rate law for the reaction NH4(aq)+ + NO2(aq)- → H2O(l) + NO2(g) is rate = k [NH4+][NO2-], and the value of k is 2.7*10-4 M-1

dybincka [34]

We are given the rate law, so we can substitute the given values for the rate constant and the concentrations of the reactants to solve for the rate of reaction. Since rate = k [NH4+][NO2-]:
rate = (2.7 x 10^-4 / M-s)(0.050 M)(0.25 M) = 3.375 x 10^-6 M/s

8 0

3 years ago

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