Pls help I’m in a test plssss

The vapor pressure of water is 23.76 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in water is sucrose. Calculate

MrRissso [65]

Answer : The vapor pressure of solution is 23.67 mmHg.

Solution:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent (water) = 23.76 mmHg

$p_s$ = vapor pressure of solution= ?

$w_2$ = mass of solute (sucrose) = 12.25 g

$w_1$ = mass of solvent (water) = 176.3 g

$M_1$ = molar mass of solvent (water) = 18.02 g/mole

$M_2$ = molar mass of solute (sucrose) = 342.3 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{23.76-p_s}{23.76}=\frac{12.25\times 18.02}{176.3\times 342.3}$

$p_s=23.67mmHg$

Therefore, the vapor pressure of solution is 23.67 mmHg.

7 0

3 years ago

For the following reaction, 9.30 grams of glucose (C6H12O6) are allowed to react with 13.8 grams of oxygen gas. glucose (C6H12O6

amid [387]

Answer:

13.7 g of CO₂

Limiting reactant: C₆H₁₂O₆

3.81 g of O₂

Explanation:

We convert the mass of the reactants to moles, in order to find out the limiting reactant and the excess reagent

9.30 g / 180 g/mol = 0.052 moles of glucose

13.8 g / 32 g/mol = 0.431 moles of oxygen

The equation is: C₆H₁₂O₆(s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

Ratio is 1:6. Let's consider this rule of three:

1 mol of glucose reacts with 6 moles of oxygen

Then, 0.052 moles of glucose must react with (0.052 . 6) /1 = 0.312 moles

We have 0.431 moles of oxygen and we only need 0.312 moles. This means that an amount of oxygen still remains after the reaction is complete:

0.431 - 0.312 = 0.119 moles. We convert the moles to mass:

0.119 mol . 32 g / 1mol = 3.81 g

In conclussion, the limiting reactant is the glucose.

6 moles of oxygen react with 1 mol of glucose

0.431 moles of O₂ will react with (0.431 . 1) /6 = 0.072 moles of glucose

We only have 0.052 moles, so it is ok to say, that glucose is the limiting cause we do not have enough glucose.

Let's verify, the maximum amount of carbon dioxide that can be formed:

1 mol of glucose can produce 6 moles of CO₂

Therefore 0.052 moles of glucose will produce (0.052 . 6) /1 = 0.312 moles

We convert the moles to mass → 0.312 mol . 44 g /1 mol = 13.7 g

6 0

3 years ago

I would like some assistance

charle [14.2K]

Answer:

#2.

Explanation:

Look at the charges. Both are positive, therefore both are cations.

6 0

2 years ago

Both white phosphorus and red phosphorus are made up only of phosphorus atoms.Does this mean they are both pure substances? Coul

choli [55]

True, both of the red and white phosphorus is a pure substance.

Both red and white phosphorus is made of phosphor element but they have a different structure which causes them to have a different color. This phenomenon called allotropes when a chemical element has two or more different form.They are not compounds as they are only made of phosphorus. A compound should be made by at least two different elements.

4 0

3 years ago

1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)

Yuki888 [10]

Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of carbon to fore 0.3009 moles of CO₂

The maximum mass of carbon dioxide that can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

3 0

3 years ago