Why is space cold ???

The normal freezing point of a certain liquid

stepladder [879]

Answer : The molal freezing point depression constant of X is $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X= $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

$[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}$

$k_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of X is $4.12^oC/m$

4 0

4 years ago

In this experiment, you were asked to calculate your percent yield of alum assuming that you started with pure aluminum. Actuall

neonofarm [45]

Answer:

Percent yield of alum is 38.6%

Explanation:

Find attached calculations for detailed explanation.

3 0

3 years ago

a couple of pennies have a mas of 5.2 grams. Asuming that the pennies are 100% percent copper,how many moles of copper are the p

dangina [55]

Answer:

5.2g copper (Cu) => 0.082 moles copper (2 sig.figs.)

Explanation:

mole conversions:

grams to moles => divide by formula wt.

moles to grams => multiply by formula wt.

gas volumes to moles => divide volume by 22.4Liters/mole (STP conditions only)

This problem:

mass to moles => divide by formula wt.

mass = 5.2g = 5.2g/63.5g/mole = 0.082 moles copper (2 sig.figs.)

5 0

3 years ago

The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur

anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × $10^{7}$

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × $10^{7}$

ΔG° = -2494.2㏑ 1.7 × $10^{7}$ = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

Learn about formation constant

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8 0

2 years ago

Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w

quester [9]

Answer:

0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

4 0

3 years ago