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3 years ago

The dissociation of a weak electrolyte is suppressed when

Chemistry

2 answers:

Vanyuwa [196]3 years ago

7 0

A compound that undergoes only partial dissociation when dissolved in water

adelina 88 [10]3 years ago

4 0

Answer: a strong electrolyte with a different ion than the weak electrolyte is added.

Explanation:

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Calculate the value of the diffusion coefficient D (in m2/s) at 547°C for the diffusion of some species in a metal; assume that

svetlana [45]

Answer : The value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

Explanation :

Formula used :

$D=D_o\times \exp \left(-\frac{Q_d}{RT}\right )$

where,

D = diffusion coefficient = ?

$D_o$ = $5.6\times 10^{-5}m^2/s$

$Q_d$ = $177kJ/mol$

R = gas constant = 8.314 J/mol.K

T = temperature = $547^oC=273+547=820K$

Now put all the given values in the above formula, we get:

$D=5.6\times 10^{-5}m^2/s\times \exp \left(-\frac{177kJ/mol}{(8.314J/mol.K)\times (820K)}\right )$

$D=2.97\times 10^{-16}m^2/s$

Thus, the value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

6 0

3 years ago

What is the net ionic charge of calcium ion

dezoksy [38]

Explanation: Ca generally loses two electrons from its outer shell to form a Ca2+ ion.
Hope this helped!!
Could I get brainliest by chance?!

7 0

4 years ago

What is the total number of atoms in the molecule<br> H2SO4?

prisoha [69]

Answer:

Sulfuric acid (H2SO4) contains:

2 Hydrogen 1 Sulfer

4 Oxygen

Explanation:

5 0

3 years ago

Read 2 more answers

The cell membrane is found in all cells *<br><br> True<br> False

brilliants [131]

Answer:

true

Explanation:

all cells will have plasma membranes

3 0

3 years ago

A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i

Alexeev081 [22]

Answer:

$MM_{acid}=140.1g/mol$

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

$n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}$

Thus, solving for the moles of the acid, we obtain:

$n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol$

Then, by using the mass of the acid, we compute its molar mass:

$MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol$

Regards.

7 0

3 years ago