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3 years ago

Sample of gas occupies 10.0 L at 50 ˚C. Assuming that the pressure is constant what volume will

Chemistry

1 answer:

scZoUnD [109]3 years ago

8 0

Answer:

<em>20 Liters</em>

Explanation:

If the pressure is supposed to be constant, one of Charles - Gay Lussac's laws can be used to solve the exercise. His statement says that "the volume of the gas is directly proportional to its temperature at constant pressure", mathematically it would be:

$\frac{V1}{T1} = \frac{V2}{T2}$

Considering T₁ = 50 ° C; V₁ = 10.0 L; and T₂ = 100 ° C, we can calculate the volume V₂ Clearing it from the equation and replacing the values to perform the calculation:

V2= (V1 x T2) / T1 = (10.0 L x 100°C) / 50 °C = 20 L

Therefore, <em>the gas at 100 ° C will occupy a volume of 20.0 L</em>.

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3 years ago

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3 years ago

It's almost time for bed but before you're ready to go to sleep you decide to make yourself a cup of hot chocolate and a tasty s

ehidna [41]

1. Milk.

2. What is in the bowl the popcorn, cereal, and pretzels.

3. Hot chocolate and hot milk.

4. Hot chocolate mix.

5. Hot milk.

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3 years ago

The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this

azamat

Answer : The value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

Solution : Given,

Solubility of $BaCO_3$ in water = $4.4\times 10^{-5}mole/L$

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

$BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3$

Initially - 0 0

At equilibrium - s s

The Solubility product will be equal to,

$K_{sp}=[Ba^{2+}][CO^{2-}_3]$

$K_{sp}=s\times s=s^2$

$[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L$

Now put all the given values in this expression, we get the value of solubility constant.

$K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2$

Therefore, the value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

3 0

3 years ago

Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m h

Alborosie

Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L

∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = $\frac{number of moles of [OH-]}{volume of solution (L)}$ = $\frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)}$ = 0.03199 M

6 0

3 years ago