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2 years ago

How many moles is 573.28 g of AgCl? A. 3 mol B. 5 mol C. 4 mol D. 2 mol

Chemistry

1 answer:

hodyreva [135]2 years ago

7 0

Answer:

Explanation:

Just divide the mass of how much AgCl by the mass per mole to give you the number of moles, 573.28g 143.318gmol−1 AgCl = 4.00molAgCl

Hope this helps! <3

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When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago

If 3.31 moles of argon gas occupies a volume of 100 L what volume does 13.15 moles of argon occupy under the same temperature an

kumpel [21]

Answer:

397 L

Explanation:

Recall the ideal gas law:

$\displaystyle PV = nRT$

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

$\displaystyle \frac{P}{RT} = \frac{n}{V}$

The left-hand side is simply some constant. Hence, we can write that:

$\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}$

Substitute in known values:

$\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})} = \frac{(13.15\text{ mol })}{V_2}$

Solving for V₂ yields:

$\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}$

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.

(Assuming 100 L has three significant figures.)

3 0

1 year ago

Electrons are embedded in a mass of positively charged matter.

grandymaker [24]

Answer:

plum pudding model .

Explanation:

the electrons were 'like plums embedded in a pudding'. Also called the Raisin Bread Model.

5 0

2 years ago

Please help me with question 7. Thank you so much.

tigry1 [53]

Answer:

The system is not at equilibrium and the reaction will proceed to the left.

Explanation:

Step 1: Write the balanced equation

H₂(g) + CO₂(g) ⇄ CO(g) + H₂O(g)

Step 2: Calculate the reaction quotient (Q)

The reaction is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.

Q = [CO] × [H₂O] / [H₂] × [CO₂]

Q = 0.610 × 0.695 / 0.425 × 0.500 = 2.00

Since Q ≠ Kc, the reaction is not at equilibrium.

Since Q > Kc, the reaction will proceed to the left.

4 0

2 years ago

How do you convert Pa*mm^2 to Pa*m^2?

Elden [556K]

Answer:

$1 Pa\times mm^{2} =10^{-6}Pa\times m^{2}$

Explanation:

Pascal(Pa) : It is the SI unit of pressure.It is equal to 1 Newton per meter square(Pa = $N/m^{2}$ )

$Pa\times mm^{2}$ is the Unit of Force(Newton=N)

Calculation,

It is required to convert $mm^{2}$ to $m^{2}$ only, Pa is same in both so no need to alter Pa.

1 m = 1000 mm (see the image)

And,

1 mm = 0.001 m

$1 mm^{2} = 0.001\times 0.001 m^{2}$ (Squaring both side)

$1 mm^{2} = 0.000001 m^{2}$

$1 mm^{2} = 10^{-6}m^{2}$

So,

$1 Pa\times mm^{2} =10^{-6}Pa\times m^{2}$

3 0

2 years ago