Mass 16g 1 cm .5 cm 2 cm

You need 65.0 g of aluminum for an experiment. How many atoms of aluminum is that?

nikdorinn [45]

Answer:

Explanation:

To convert the mass of Aluminum to atoms, we must first convert to moles by using the formula;

mole = mass/molar mass

5 0

3 years ago

A solar eclipse will last 1 to _____ minutes.

stellarik [79]

1st qn - B
2nd qn - A
Last qn - Yes

6 0

3 years ago

Read 2 more answers

Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significa

Black_prince [1.1K]

Answer: The equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

Explanation:

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid $(H_2C_2O_4)$ is a weak acid and will dissociate 2 hydrogen ions.

The chemical equation for the first dissociation of oxalic acid follows:

$H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)$

Initial: 0.20

At eqllm: 0.20-x x x

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}$

We know that:

$Ka_1\text{ for }H_2C_2O_4=0.059$

Putting values in above equation, we get:

$0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

The chemical equation for the second dissociation of oxalic acid:

$HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)$

Initial: 0.083

At eqllm: 0.083-y 0.083+y y

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}$

We know that:

$Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}$

Putting values in above equation, we get:

$6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

7 0

3 years ago

Which of the following would diffuse through a cell membrane the most easily?

ZanzabumX [31]

You didn’t give us the options???

anyways: Water diffusion is called osmosis. Oxygen is a small molecule and it's nonpolar, so it easily passes through a cell membrane. Carbon dioxide, the byproduct of cell respiration, is small enough to readily diffuse out of a cell. Small uncharged lipid molecules can pass through the lipid innards of the membrane.

8 0

3 years ago

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WHAT MASS OF 1,1 DICHLOROEHTANE MUST BE MIXED WITH 100G OF 1,1 DICHLOROTETRAFLUOROEHTANE TO GIVE A SOLUTION WITH VAPOR PRESSURE

vaieri [72.5K]

This is an incomplete question.

The complete question is:

1,1-dichlorotetrafluoroethane, CF3CCL2F, has a vapor pressure of 228 torr. What mass of 1,1-dichloroethane must be mixed with 100.0 g of 1,1-dichlorotetrafluoroethane to give a solution with vapor pressure 157 torr at 25 degrees celsius?

Answer: 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at $25^0C$

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute

= $\frac{\text {moles of solute}}{\text {total moles}}$

Given : x g of solute is present in 100 g of solvent

moles of solute (1,1 DICHLOROEHTANE) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{xg}{98.96g/mol}moles$

moles of solvent (1,1 DICHLOROTETRAFLUOROEHTANE ) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{100g}{170.92g/mol}=0.58moles$

Total moles = moles of solute + moles of solvent = $\frac{xg}{98.96g/mol}+0.58$

$x_2$ = mole fraction of solute = $\frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$\frac{228-157}{157}=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$0.45=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$x=46.9g$

Thus 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at $25^0C$

5 0

4 years ago