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3 years ago

!!MARKING BRAINLIEST!! Please help this is due at midnight!! No websites please just help me! :(

Chemistry

1 answer:

9966 [12]3 years ago

6 0

Answer:

I'll do the first one for you. The reason why I'm not going to do the rest is because this is pretty simple stuff. I'll explain how I got the answer, please read it ^^ the rest of the problems should be a breeze.

1. 5.454285714285714 liters, or approx. 5.45 liters

Explanation:

P1V2 = P2V2

P1 refers to the original pressure. V1 refers to the original volume, or the amount of space the gas takes up.

P2 and V2 refer to the final pressure or volume, accordingly.

You insert the values into the equation, like so:

(8.3)(46) = (70)(x)

Now, multiply.

381.8 = 70x

Use inverse operations to find the value of x. Divide 381.8 by 70 to isolate x.

381.8/70 = x

5.454285714285714 = x

The volume of the gas when the pressure is increased to 70.0 mm Hg is approximately 5.45 liters. Don't forget about the units at the end, when you write your final answer.

Important! When pressure increases, volume decreases, and vice versa. Volume and pressure for gases are inversely proportional. So even though the pressure increased, that doesn't mean the volume increases, too.

You can check your answers easily!

Just multiply your final answer by its corresponding pressure or volume and compare it to the other. I hope that made sense. Like so:

5.454285714285714 x 70 = 381.8

8.3 x 46 = 381.8

That makes P1V2 DOES equal P2V2, and your answer is correct.

I hope this helped in time for you to submit it before the deadline! Good luck.

Tips!

For #2: I'm pretty sure the mentioning of the temperature (25.0 °C) doesn't matter. You can ignore it, it won't affect your calculations.

For #4: the standard pressure in mm Hg (millimeters of mercury) is 760 mm Hg. That's your P2.

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2NaOH+CO2----->Na2CO3+H2O

kondaur [170]

Answer:

13.2 g Na2CO3

Explanation:

Convert 10.0 g NaOH to mol.

10.0 g x 1 mol/39.997 g = 0.250 mol

Use mol ration given by the equation: 2 mol NaOH to 1 mol Na2CO3

0.250 mol NaOH x 1 mol Na2CO3/2 mol NaOH = 0.125 mol Na2CO3

Finally, convert the moles of Na2CO3 to grams.

0.125 mol Na2CO3 x 105.99 g/1 mol = 13.2 g

8 0

3 years ago

A generic salt, AB 2 , has a molar mass of 345 g/mol and a solubility of 8.70 g/L at 25 °C. AB 2 (s) − ⇀ ↽ − A 2 + (aq) + 2B − (

Brrunno [24]

Answer : The value of $K_{sp}$ of the generic salt is, $1.60\times 10^{-5}$

Explanation :

As we are given that, a solubility of salt is, 8.70 g/L that means 8.70 grams of salt present in 1 L of solution.

First we have to calculate the moles of salt $(AB_2)$

$\text{Moles of }AB_2=\frac{\text{Mass of }AB_2}{\text{Molar mass of }AB_2}$

Molar mass of $AB_2$ = 345 g/mol

$\text{Moles of }AB_2=\frac{8.70g}{345g/mol}=0.0252mol$

Now we have to calculate the concentration of $A^{2+}\text{ and }B^-$

The equilibrium chemical reaction will be:

$AB_2(s)\rightleftharpoons A^{2+}(aq)+2B^-(aq)$

Concentration of $A^{2+}$ = $\frac{0.0252mol}{1L}=0.0252M$

Concentration of $B^-$ = $\frac{0.0252mol}{1L}=0.0252M$

The solubility constant expression for this reaction is:

$K_{sp}=[A^{2+}][B^-]^2$

Now put all the given values in this expression, we get:

$K_{sp}=(0.0252M)\times (0.0252M)^2$

$K_{sp}=1.60\times 10^{-5}$

Thus, the value of $K_{sp}$ of the generic salt is, $1.60\times 10^{-5}$

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3 years ago

Ionic solids dissolve in water and break up into their ions. However, some ionic solids only partially dissolve, leaving a signi

Xelga [282]

Answer: The expression of $K_{sp}$ for calcium fluoride is $K_{sp}=[Ca^{2+}][2F^-]^2$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The given chemical equation follows:

$CaF_2\rightleftharpoons Ca^{2+}(aq.)+2F^-(aq.)$

1 mole of calcium fluoride produces 1 mole of calcium ions and 2 moles of fluorine ions.

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=[Ca^{2+}][2F^-]^2$

Hence, the expression of $K_{sp}$ for calcium fluoride is $K_{sp}=[Ca^{2+}][2F^-]^2$

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3 years ago

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2 years ago

A chemistry student adds a quantity of an unknown solid compound X to 750. mL of distilled water at 26.° C. After 10 minutes of

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Explanation:

The given data is as follows.

X = 12 g, Volume of water = 2 L,

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$\frac{12 g}{2 L}$

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Solubility = $\frac{6 g}{1000 ml}$

= 0.006 g/ml

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3 years ago