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Gelneren [198K]
2 years ago
13

A solution of HCl with a volume of 25.00 mL is titrated to the endpoint, with 0.250 M

Chemistry
1 answer:
kirill [66]2 years ago
8 0

Answer:

0.3456\ \text{M}

Explanation:

V_1 = Volume of NaOH = 34.56 mL

V_2 = Volume of HCl = 25 mL

M_1 = Concentration of NaOH = 0.25 M

M_2 = Concentration of HCl

When endpoint is reached the number of moles of NaOH will be equal to the number of moles of HCl

M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{0.25\times 34.56}{25}\\\Rightarrow M_2=0.3456\ \text{M}

Concentration of HCl is 0.3456\ \text{M}.

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a 1.00litre vessel contains 0.215 mole of nitrogen gas and 0.0118 mole of hydrogen gas at 25°C. determine the partial pressure o
stepladder [879]

Answer:

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8 0
2 years ago
He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
Xelga [282]

Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

8 0
3 years ago
Read 2 more answers
At what temperature, would the volume of a gas
PtichkaEL [24]

Explanation:

P1V1 = nRT1

P2V2 = nRT2

Divide one by the other:

P1V1/P2V2 = nRT1/nRT2

From which:

P1V1/P2V2 = T1/T2

(Or P1V1 = P2V2 under isothermal conditions)

Inverting and isolating T2 (final temp)

(P2V2/P1V1)T1 = T2 (Temp in K).

Now P1/P2 = 1

V1/V2 = 1/2

T1 = 273 K, the initial temp.

Therefore, inserting these values into above:

2 x 273 K = T2 = 546 K, or 273 C.

Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.

From the ideal gas equation:

V = nRT/P or at constant pressure:

V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.

8 0
2 years ago
Whats Potassium bond type? Please help!
olga_2 [115]

Answer:

A metallic bond.

Explanation:

Potassium is a metal (alkali metal), hence its bonds are metallic bonds.

Hope this helped!

5 0
3 years ago
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Please help me with this question
vlada-n [284]
The answer to that question is B
7 0
2 years ago
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