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3 years ago

Can someone pls help me with this question?:

Chemistry

1 answer:

dangina [55]3 years ago

6 0

Answer:

James Chadwick

Explanation:

In his first experiment, Rutherford was unable to predict the total mass of nucleus. He noticed that if only electrons and protons are present in the nucleus then by adding their masses, it falls short of the net mass of the atom. so he concluded that there must be another particle and that Particle was later discovered by James Chadwick in 1932.

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What is the correct answer ??

Ber [7]

Answer:

[H_3 O^+] = 1.0 ×10^-13

Explanation:

If we multiply the left side we get, 1e - 13. We add 13 to the right side while subtracting the remaining value from the left side (H3O+) than combine like terms. As you will get pH = 13.00

7 0

3 years ago

Read 2 more answers

Enter a balanced equation for the complete combustion of liquid C3H7OH. Express your answer as a chemical equation. Identify all

Vlad1618 [11]

2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)

Explanation:

To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.

Bellow we have the balanced chemical equation of the complete combustion of C₃H₇OH:

C₃H₇OH (l) + (9/2) O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)

to have integer coefficients we multiply the reaction with 2:

2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)

where:

(l) - liquid

(g) - gaseous

Learn more about:

combustion reaction

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7 0

3 years ago

A certain gas is present in a 13.0 l cylinder at 4.0 atm pressure. if the pressure is increased to 8.0 atm , the volume of the g

ella [17]

Boyle's Law is k = PV so Initial k = 13.0 L x 4.0 atm = 52 L atm Final kf = 6.5 L x 8 atm = 52 L atm The gas obeys Boyle's Law The answer with two significant figures separated by a comma is k = 52, kf = 52.

5 0

3 years ago

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A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What i

spayn [35]

Answer:

0.681 atm

Explanation:

To solve this problem, we make use of the General gas equation.

Given:

P1 = 785 torr

V1 = 2L

T1 = 37= 37 + 273.15 = 310.15K

P2 = ?

V2 = 3.24L

T2 = 58 = 58+273.15 = 331.15K

P1V1/T1 = P2V2/T2

Now, making P2 the subject of the formula,

P2 = P1V1T2/T1V2

P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

P2 = 515.715 Torr

We convert this to atm: 1 torr = 0.00132 atm

515.715 Torr = 515.715 * 0.00132 = 0.681 atm

8 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