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3 years ago

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The degree to which various compounds will dissociate in solution varies greatly.

1 answer:

uranmaximum [27]3 years ago

5 0

Answer: True

Explanation:

Weak electrolytes are those solutions which do not undergo complete dissociation when dissolved in water. The dissociation of weak electrolytes is given by an equilibrium.

Example: $CH_3COOH\rightleftharpoons CH_3COO^-+H^+$

Strong electrolytes are those solutions which undergo complete dissociation when dissolved in water. The dissociation of strong electrolytes is given by a right arrow.

Example: $HCl\rightarrow H^++Cl^-$

Thus the degree to which various compounds will dissociate in solution varies greatly is true.

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Calculate the volume 3.00 moles of a gas will occupy at 24.0˚C and 1.003 atm. *

vitfil [10]

Answer: 72.93 litres

Explanation:

Given that:

Volume of gas (V) = ?

Temperature (T) = 24.0°C

Convert 24.0°C to Kelvin by adding 273

(24.0°C + 273 = 297K)

Pressure (P) = 1.003 atm

Number of moles (n) = 3 moles

Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1

Then, apply ideal gas equation

pV = nRT

1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K

1.003 atm•V = 73.15 atm•L

Divide both sides by 1.003 atm

1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm

V = 72.93 L

Thus, the volume of the gas is 72.93 litres

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3 years ago

A piece of tin has a mass of 16.52 g and a volume of 2.26 cm2 what is the density of tin?

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M/V=D
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2 years ago

State the differences between Transition metals and S-blocks metals<br>

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2 years ago

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Read 2 more answers

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

2 years ago