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2 years ago

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A NaOH (aq) stock solution was created by dissolving 3.88 g NaOH in water to create a 100.00 mL solution. What is the concentrat

ion of this solution

1 answer:

Ierofanga [76]2 years ago

8 0

Answer:

I have don 3 so far

Explanation:

Brainlyis the best

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calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.

Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

$P_w=P_p+P_s$

$P_w=P_p+(-iCRT)$

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= $P_w$

$P_p=0 bar$ (At standard temperature)

$P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar$

$P_w=0 bar+(-3.406 ) bar$

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

7 0

2 years ago

III. For the following four sections: If you are given the formula, name the

serious [3.7K]

Answer:

15.Potassium oxide

16.Calcium chloride

17.Aluminium sulphide

18.CaS

Explanation:

15.K is the chemical symbol of Potassium and generally the name of the non-metal at the end of a formula has the suffix '-ide' and since O is oxygen, the name becomes Potassium oxide.

16. The same applies here. Ca is Calcium and Cl is Chlorine but since its the non-metal at the end, it ends in -ide. So Calcium chloride.

17.The same applies here too. Al is Aluminium and S is Sulphur so Aluminium sulphide.

18. Calcium's symbol is Ca and that of Sulphur is S and that gives the formula CaS.

5 0

3 years ago

A 687.80-g sample of a noble is held in a 15 L cylinder tank at 3800 torr and a temperature of 22 degrees C. Identify the gas.

BARSIC [14]

Answer:.......................

8 0

2 years ago

Molecular formula for C4H10

Gnesinka [82]

I think this is what you mean:

H H H H
H-C-C-C-C-H
H H H H

OR

<span>CH3CH2CH2CH3
</span>
If not, clarify and I will be happy to help.

6 0

3 years ago

A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under

sukhopar [10]

Answer:

$\rho _2=0.22g/L$

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

$PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT$

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

$\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}$

It means we can compute the final density as shown below:

$\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}$

Now, we plug in to obtain:

$\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L$

Regards!

8 0

2 years ago