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adelina 88 [10]

2 years ago

14

A 0.10 L solution with 1.85 g of Ca(OH)2 is of what concentration, in molarity?

1 answer:

Mandarinka [93]2 years ago

3 0

Answer:

0.25M

Explanation:

The formula of concentration is c=m/M•V

c=1.85g/74g.mol•0.10L

c=0.25 M

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Calculate the boiling point of a 3.60 m aqueous sucrose solution. express the boiling point in degrees celsius to five significa

kvv77 [185]

We can use this equation for boiling point elevation:
ΔT(b) = i K(b) M
when Δ T(b) is the increase of boiling point of the solution.
and i is ( vant Hoff factor, the number of particles or ions per mole-clue.
and K(b) is boiling point increase constant for the solution ( and for water it is equal 0.52 C° Kg/mol)
We can assume i (vant Hoff factor ) = 1 as the sucrose is nonelectrolyte (not readily ionize).
So for water: Tb° = 100 c° and Kb = 0.52 c° Kg / mol
By substitute at:
ΔTb = i Kb M
∴ = 1 * 0.52 * 3.60 = 1.8432 C°
and when Tb = Tb° + ΔTb
∴ Tb = 100 + 1.8432 = 101.8432 C°

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2 years ago

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Mercury bc it is melting and getting closer

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3 years ago

Which may be true for a substance that absorbs energy?

irinina [24]

It has different moleculs sorry I am guessing

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2 years ago

Calculate the density of CO2 in g/cm3 at room temperature(25 degrees Celsuis) and pressure(1 atm) assuming it acts as an ideal g

Readme [11.4K]

Answer:

$density=1.8x10^{-3}g/mL$

Explanation:

Hello,

Considering the ideal equation of state:

$PV=nRT$

The moles are defined in terms of mass as follows:

$n=\frac{m}{M}$

Whereas $M$ the gas' molar mass, thus:

$PV=\frac{mRT}{M}$

Now, since the density is defined as the quotient between the mass and the volume, we get:

$P=\frac{m}{V} \frac{RT}{M}$

Solving for $m/V$ :

$density= m/V=\frac{PM}{RT}$

Thus, the result is given by:

$density=\frac{(1atm)(44g/mol)}{[0.082atm*L/(mol*K)]*298.15K} \\density=1.8g/L=1.8x10^{-3}g/mL$

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2 years ago

The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to

tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of $CuFeS_2$ = 183.511 g/mole

First we have to calculate the moles of Cu.

$\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles$

The moles of Cu = 4.7209 moles

From the given chemical formula, $CuFeS_2$ we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of $CuFeS_2$ = 4.4209 moles

Now we have to calculate the mass of $CuFeS_2$ .

Mass of $CuFeS_2$ = Moles of $CuFeS_2$ × Molar mass of $CuFeS_2$ = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of $CuFeS_2$ = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

3 0

2 years ago