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2 years ago

It is a very easy matter. I want the solution please

Chemistry

1 answer:

Helga [31]2 years ago

4 0

Answer:

0.125

Explanation:

For 250 ml it is 0.5 moles. If you know the Molecular weight of NaCl you multiply with 0.5 to get the answer. Required number of moles =2⋅0.25=0.5 moles.

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Which is less likely to be a relatable source of information the webpage of a university or the webpage a scientist who is tryin

faltersainse [42]

Answer:

The web page of a university

Explanation:

A scientist can be more biased within coming to information about pretty much anything. I have had multiple science teachers who seem more biased on to something else and pretend that they're right just cause they know what they are doing.

Then the university would be a great choice because its controlled by a higher state, then also the consistency of being updated.

7 0

3 years ago

When the pressure that a gas exerts on a sealed container changes from 1100 bar to 75.5 bar, the temperature changes from k to 2

Serjik [45]

Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant
$\frac{P1}{T1} = \frac{P2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation
$\frac{1100 bar}{T} = \frac{75.5 bar}{298 K}$
T = 4342 K
initial temperature was 4342 K

7 0

3 years ago

What is the ph of a solution labeled .30 trimethylamine k for trimethylamine is 7.42 x 10^4

valentina_108 [34]

Answer:the pH is 12

Explanation:

First We need to understand the structure of trimethylamine

Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline

C3H9N+ H2O --> C3H9NH + OH-

$k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}$

Then:

The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine

[OH-]= $\sqrt{ 0.3*7.42x10^{-4}}$

[OH-]=0.01

pH=14-(-log[OH-])

pH=12

5 0

3 years ago

Write the balanced chemical Equation for the following when they are heated

ale4655 [162]

Answer:

i. Lead nitrate:

2Pb (NO3)2 Δ= 2PbO+4NO2+O2

ii. Potassium chlorate:

2KClO3 → 2KCl + 3O2↑

iii. clacium carbonate:

CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

iv. cupric carbonate :

CuCO3 → CuO + CO2↑

Hope this helped

All the best!!

4 0

3 years ago

One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expa

Verdich [7]

Answer:

a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K

b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K

c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J

d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K

Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

∴ n = 1 mole

∴ PV = RTn....ideal gas

∴ P1 = 10 atm

∴ R = 0.082 atm.L/K.mol

∴ T = 300 K = T2

∴ V2 = 3*V1

⇒ W = 0.....expands freely into vacuum

⇒ ΔU = Q = 0....first law

⇒ ΔS = - nR Ln(P2/P1).....ideal gas

∴ V1*P1/T1 = V2*P2/T2

∴ T1 = T2 = 300 K

⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm

⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)

⇒ ΔS = 0.09 atm.L/K

∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0

b) heated reversibly at constant volume:

⇒ W = 0 ...at constant volume

∴ T2 = 400 K; T1 = 300 K

∴ V1 = V2

⇒ Q = ΔU = CvΔT....first law

∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas

∴ ΔT = 400 - 300 = 100 K

⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J

∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J

∴ ΔS = - nR Ln (P2/P1)

∴ P2/T2 = P1/T1...constant volume

∴ P1 = 3.33 atm

⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm

⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)

⇒ ΔS = - 0.0235 atm.L/K

c) reversibly expanded at constant temperature:

∴ T1 = T2 = 400K

∴ V2 = 3*V1

∴ ΔU = 0...constant temperature

⇒ Q = - W....fisrt law

∴ W = - ∫ PdV..... reversibly expansion

∴ P = nRT/V... ideal gas

⇒ W = - nRT ∫ dV/V

⇒ W = - nRT Ln (V2/V1)

⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)

⇒ W = - 9.134 J/K *400K = - 3653.545 J

⇒ Q = - W = 3653.545 J

⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0

∴ ΔS = - nR Ln(P2/P1)

∴ P1 = 4.44 atm

⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)

⇒ P2 = 4.44atm/3 = 1.48 atm

⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)

⇒ ΔS = -9.134J/K * 400K = - 3653.545 J

d) reversibly cooled at constant pressure:

∴ T2 = 300 K; T1 = 400 K

∴ P2 = P1

⇒ Q = ΔH = CpΔT

∴ Cp = 20.8 J/K.mol

∴ ΔT = 300 - 400 = - 100 K

⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J

⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J

⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J

∴ ΔS = ∫ δQ/T = ∫ nCpdT/T

⇒ ΔS = nCp Ln (T2/T1)

⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K

7 0

3 years ago