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3 years ago

11

Shown here is a person shaving. Under magnification, the shaving foam might look like the image above the shaver. What type of m

ixture does the foam demonstrate? Give your reasoning.

1 answer:

SOVA2 [1]3 years ago

5 0

Include in your answer:
-foam is a colloid.
-colloids include gas dispersed in a liquid.
-colloids include gas dispersed in a solid.

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3. Calculate the answers to the appropriate number of significant figures. e) 43.678 x 64.1 = f) 1.678/0.42 =

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Where’s the pictures

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3 years ago

How would you prepare 3.5 L of a 0.9M solution of KCl?

PolarNik [594]

$V=3,5L\\
Cm=0,9M\\
M_{KCl}=74\frac{g}{mol}\\\\
C_{m}=\frac{n}{V}\\\\
n=\frac{m}{M}\\\\
C_{m}=\frac{m}{MV} \ \ \ \Rightarrow \ \ \ m=C_{m}MV\\\\
m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g$

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.

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3 years ago

A sample of helium gas has a pressure of 1.20 atm at 22°c. at what celsius temperature will the helium reach a pressure of 2.00

sukhopar [10]

Gay-Lussac's law gives the relationship between pressure and temperature of a gas.
it states that for a fixed amount of gas of constant volume pressure is directly proportional to temperature.
P/T = k
where P - pressure, T - temperature and k - constant
$\frac{P1}{T1} = \frac{P2}{T2}$
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
temperature should be in the kelvin scale,
T1 = 22 °C + 273 = 295 K
substituting the values in the equation
$\frac{1.20atm}{295 K} = \frac{2.00 atm}{T}$
T = 492 K
new temperature - 492 - 273 = 219 °C

5 0

3 years ago

Which of the following is from smallest to largest

aliya0001 [1]

Atom, molecule , organelle, cell , tissue organ, organ systems, organism population, community, ecosystem, biome, biosphere

3 0

3 years ago

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0

3 years ago