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3 years ago

12

a solution of prepared by dissolving 20.0 brams of mavni fluride (MgF2) in 80.0 of water (H2O). whag is the mole fraction of mag

nesium fluride in the solution?

1 answer:

Serjik [45]3 years ago

3 0

Answer:

2000.1234 trebuie sa fie asa

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An increase in the surface area of reactants in a heterogeneous reaction will result in what?

marusya05 [52]

It will result in an increase in the rate of rxn

4 0

3 years ago

What is the molality of a solution if 100.0 g of glucose (C&Hi20e) were dissolved into 750. mL of water?

hodyreva [135]

<u>Answer:</u> The molality of solution is 0.740 m.

<u>Explanation:</u>

To calculate the mass of solvent (water), we use the equation:

$Density=\frac{Mass}{Volume}$

Volume of water = 750 mL

Density of water = 1 g/mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{750mL}\\\\\text{Mass of water}=750g$

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(C_6H_{12}O_6)$ = 100.0 g

$M_{solute}$ = Molar mass of solute $(C_6H_{12}O_6)$ = 180 g/mol

$W_{solvent}$ = Mass of solvent (water) = 750 g

Putting values in above equation, we get:

$\text{Molality of }C_6H_{12}O_6=\frac{100\times 1000}{180\times 750}\\\\\text{Molality of }C_6H_{12}O_6=0.740m$

Hence, the molality of solution is 0.740 m.

6 0

3 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$

The pressure is constant, so: $w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}$ (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

$U_{2}-U_{1}=500-(3*75)=275kJ$

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

$P_{1}V_{1}=nRT_{1}$

$P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}$

Subtracting the first from the second:

$1.5P_{1}V_{1}=nR(T_{2}-T_{1})$

Isolating $T_{2}$ :

$T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}$

Assuming that it is water steam, n=0.1666 kmol

$V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}$

$T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76$ ºC

7 0

3 years ago

10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of

Ilia_Sergeevich [38]

Answer:

0.355 N of HF

Explanation:

The titration reaction of HF with KOH is:

HF + KOH → H₂O + KF

<em>Where 1 mole of HF reacts per mole of KOH</em>

<em />

Moles of KOH are:

0.0296L × (0.120 equivalents / L) = 3.552x10⁻³ equivalents of KOH = equivalents of HF.

As volume of the titrated solution was 10.0mL, normality of HF solution is:

3.552x10⁻³ equivalents of HF / 0.010L =<em> 0.355 N of HF</em>

3 0

3 years ago

A gas has a volume of 1000.0 mL at a temperature of 20.OK and a pressure

yanalaym [24]

Answer:

4000mL

Explanation:

Using the combined gas law equation as follows:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (mL)

V2 = final volume (mL)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information given in this question:

V1 = 1000mL

T1 = 20K

P1 = 1.0atm

V2 = ?

P2 = 0.5atm

T2 = 40K

Using P1V1/T1 = P2V2/T2

1 × 1000/20 = 0.5 × V2/40

1000/20 = 0.5V2/40

50 = 0.5V2/40

50 × 40 = 0.5V2

2000 = 0.5V2

V2 = 2000/0.5

V2 = 4000mL

4 0

3 years ago