Consider the partial equation.

A 1.00 L of a solution is prepared by dissolving 125.6 g of NaF in it. What would be the molarity of this solution?

DedPeter [7]

Answer:

2.99 M

Explanation:

In order to solve this problem we need to keep in mind the definition of molarity:

Molarity = moles of solute / liters of solution

In order to calculate the moles of solute, we convert 125.6 g of NaF into moles using its molar mass:

125.6 g NaF ÷ 42 g/mol = 2.99 mol NaF

As the volume is already given, we can proceed to calculate the molarity:

Molarity = 2.99 mol / 1.00 L = 2.99 M

4 0

2 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago

What is the solubility of a solid in a solvent?

patriot [66]

Answer:

Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. ... The solvent is often a solid, which can be a pure substance or a mixture.

Explanation:

5 0

3 years ago

statement 1,mass is conserved durning the reaction. statement 2 sum of mass and energy is conserved durning the reaction. which

maksim [4K]

Answer:

Explanation:

None of the statement is true for both chemical and nuclear reactions. In chemical reactions, mass is always conserved and the type of atoms are also conserved.

6 0

3 years ago

You need to prepare 150 mL of 0.1 M solution of silver chloride. How much silver chloride is required?

Rudiy27

Answer:

amount of silver chloride required is 0.015 moles or 2.1504 g

Explanation:

0.1M AgCL means 0.1mol/dm³ or 0.1mol/L

1L = 1000mL

if 0.1mol of AgCl is contained in 1000mL of solution

then x will be contained in 150mL of solution

cross multiply to find x

x = (0.1*150)/1000

x= 0.015 moles

moles of silver chloride present in 150 mL of solution is 0.15 moles

To convert this to grams, simply multiply this value by the molar mass of silver chloride

molar mass of silver chloride AgCl =107.86 + 35.5

=143.36 g/mol

mass of AgCl = moles *molar mass

=0.015*143.36

=2.1504g

=

4 0

3 years ago