Sedimentation increases with an increase in land used for agriculture.

Which of the following ion is the least and the most hydrated? OH- , Li+, K+, Na+

vodomira [7]

I don’t know either lol

5 0

3 years ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

3 0

3 years ago

What is the molarity of a solution in which 7.1 g of sodium sulfate is dissolved in enough water to make 100. mL of solution?

Sidana [21]

Answer:

0.50 M

Explanation:

Given data

Mass of sodium sulfate (solute): 7.1 g

Volume of solution: 100 mL

Step 1: Calculate the moles of the solute

The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

$7.1g \times \frac{1mol}{142.02g} = 0.050mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

$100mL \times \frac{1L}{1000mL} =0.100L$

Step 3: Calculate the molarity of the solution

$M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M$

3 0

3 years ago

Read 2 more answers

Which substance in the table below has the strongest intermolecular forces

tatiyna

Answer : Option A) HF.

Explanation : Assuming the complete question as per the attachment.

H-F bonds are formed because of hydrogen bonding between H atom and F atom and is considered to be strongest amongst the rest other options given.

As given in the table it is clear that the molar mass of HF molecule is the lowest. It shows that, when dispersion increases in the molecule molecular weight also increases.

7 0

4 years ago

Read 2 more answers

If 15 grams of Carbon dioxide is produced in a chemical reaction, how many grams of Carbon must be consumed in the reaction if w

irinina [24]

Answer:

4.13 g

Explanation:

Data Given:

Amount of CO₂ Produced = 15 g

Amount of Oxygen = 11 g

Amount of Carbon used = ?

Solution:

Suppose Carbon dioxide (CO₂) is formed by the reaction of carbon and oxygen then the reaction will be as below

C + O₂ -------------> CO₂

1 mol 1 mol 1 mol

we come to know from the above reaction that

1 mole of carbon react with 1 mole of oxygen to produce 1 mol of carbon dioxide.

molar mass of C = 12 g/mol

molar mass of O₂ = 32 g/mol

molar mass of CO₂ = 12 + 2(16) = 44 g/mol

if we represent mole in grams then

C + O₂ -------------> CO 1 mol (12 g/mol) 1 mol (32 g/mol) 1 mol (44 g/mol)

C + O₂ -------------> CO₂

12 g 32 g 44 g

So,

we come to know that 32 g of Oxygen combine with 12 g of oxygen produce 44 g CO₂

So now how much of Carbon will be combine with 11 g of oxygen

apply unity formula

32 g of O₂ ≅ 12 g of C

11 g of O₂ ≅ g of C

by doing cross multiplication

g of C = 12 g x 11 g / 32 g

g of C = 132 g / 32 g

g of C = 4.13 g

So,

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

to check this answer

we use the above information

12 g of C ≅ 44 g of CO₂

4.13 g of C ≅ g of CO₂

by doing cross multiplication

g of CO₂ = 44 g x 4.13 g / 12 g

g of CO₂ = 15g

So it is confirmed that

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

4 0

3 years ago