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3 years ago

10

Is mass conserved when 200 g of water undergoes a physical change? Use complete sentences to support your answer by explaining h

ow this can be demonstrated.
help pleaseeee
its science

1 answer:

andrew-mc [135]3 years ago

7 0

Answer:

-200

Explanation:

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What is a valence electron and why are they so important to a chemist?

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The number of dots around an element symbol in a Lewis dot structure equals the total number of electrons in an atom of that ele

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3 years ago

How reliable is the perfect gas law in comparison with the van der waals equation?calculate the difference in pressure of 10.00g

Semmy [17]

Answer:

The perfect gas law is reliable at the limit of pressure tending to zero. That is to say, at low pressure. The Van der Waals law instead can be used at higher pressures.

Explanation:

The perfect law equation is

$P.V = n.R.T$

The perfect gas law considers that the gas molecules have no interaction between each other and each molecule has null volume. This condition happens at the limit of pressure tending to zero.

The Van der Waals equation is

$(P +\frac{n^{2}.a}{V^{2}})(V-n.b) = n.R.T$

On the other hand, the Van der Waals law has two extra terms, one to consider the interaction of the molecules (n2a/V2) and other to consider the volume of the molecules (V-nb). These terms make a better approximation to a real gas.

Using the two equations to calculate the pressure (P) for CO2 with

<em>V = 100cm3 = 0.1L</em>

<em>T = 25ºC = 298K</em>

<em>n = 10g/(44g/mol) = 0.23mol</em>

<em>a = 3.658 atm.L2/mol2</em>

<em>b = 0.0429 L/mol</em>

With the perfect law equation:

$P = \frac{n.R.T}{V} = \frac{0.23molx0.082\frac{atm.L}{K.mol}x298K }{0.1L} = 56.2atm$

With the Van del Waals law:

$P = \frac{n.R.T}{V-n.b} -\frac{n^{2}.a}{V^{2} }$

$P = \frac{0.23molx0.082\frac{atm.L}{K.mol}x298K}{0.1L - 0.23molx0.0429\frac{L}{mol} }-\frac{(0.23mol)^{2}x3.658\frac{atm.L^{2}}{mol^{2}}}{(0.1L)^{2}} = 43 atm$

At this case the difference between the results of the two equations is due to the big mass of CO2 which produce a high pressure.

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3 years ago

A sodium hydroxide solution that contains 24.8 grams of NaOH per L of solution has a density of 1.15 g/mL. Calculate the molalit

poizon [28]

<u>Answer:</u> The molality of NaOH in the solution is 0.551 m

<u>Explanation:</u>

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.15 g/mL

Volume of solution = 1 L = 1000 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.15 g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.15g/mL\times 1000mL)=1150g$

We are given:

Mass of solute (NaOH) = 24.8 grams

Mass of solution = 1150 grams

Mass of solvent = Mass of solution - mass of solute = [1150 - 24.8] g = 1125.2 g

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

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

Where,

$m_{solute}$ = Given mass of solute (NaOH) = 24.8 g

$M_{solute}$ = Molar mass of solute (NaOH) = 40 g/mol

$W_{solvent}$ = Mass of solvent = 1125.2 g

Putting values in above equation, we get:

$\text{Molality of }NaOH=\frac{24.8\times 1000}{40\times 1125.2}\\\\\text{Molality of }NaOH=0.551m$

Hence, the molality of NaOH in the solution is 0.551 m

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