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2 years ago

1. Why is the box sitting on the table not moving?

Chemistry

1 answer:

Rashid [163]2 years ago

7 0

I believe it’s a because something cannot move unless a force is acted upon it

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Use the periodic table to answer this question. Decomposing calcium carbonate yields calcium oxide and carbon dioxide. What info

Anuta_ua [19.1K]

Atomic number and place on table.

8 0

3 years ago

Read 2 more answers

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

2 years ago

the reversible reaction N2(g) 3H2(g) <--> 2NH3(g) produces ammonia, which is a fertilizer. At equilibrium, a 1L flask cont

Alik [6]

Answer:

Explanation:

the chemical equilibrium constant can be easily calculated since the concentrations at equilibrium are given.the calculation shows the value of Kc for the reversible reaction and forward reaction

Download pdf

7 0

2 years ago

Calculate the pOH of a solution if the concentration of hydroxide ions (OH-) is 1.9 x 10-5M?

OLEGan [10]

Answer:

9.28

Explanation:

pOH refers to a measure of hydroxide ions concentration. pOH tells about the alkalinity of a solution. If pOH is less than 7 then aqueous solutions are alkaline, acidic if pOH is greater than 7 and neutral if pOH is equal to 7.

Concentration of the hydroxide ions = 1.9 x 10-5 M

pH = $-log(1.9\times 10^{-5})=4.72$

pOH = 14 - pH

=14 - 4.72 = 9.28

6 0

3 years ago

The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for

jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

KP = 0.13 = $\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}$

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

XSO₂ = 0.58/3.29 = 0.176

XO₂ = 1.29/3.29 = 0.392

XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

p(SO₂) = 0.176 * PT

p(O₂) = 0.392 * PT

p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

$\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm$

5 0

3 years ago