compare and contrast the benefits with the potential negative effects of utilizing hydraulic fracturing to produce oil.

How many moles of glucose (C6H12O6) are in 1.6 liters of a 0.75 M C6H12O6 solution? show all work

PtichkaEL [24]

Moles glucose =1.6 L× 0.75 mol/Let =1.2

4 0

3 years ago

Read 2 more answers

The [H3O+] of a solution is 2.5 x10-7. What is the pOH of the solution?

xz_007 [3.2K]

Answer:

7.4.

Explanation:

∵ pH = -log[H₃O⁺].

∴ pH = -log(2.5 x 10⁻⁷) = 6.6.

∵ pH + pOH = 14.

∴ pOH = 14.0 - pH = 14.0 - 6.6 = 7.4.

6 0

4 years ago

Help please. ASAP

Stella [2.4K]

Answer:

COol

Explanation:

Answer A

7 0

3 years ago

A bottle contains 3.100 ml of a liquid. the total mass of the bottle and the liquid together is 6.300 g. the mass of the empty b

Olenka [21]

The answer is a. 0.665 g/m

7 0

4 years ago

Read 2 more answers

A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/gmL. The student notices that the vo

Kruka [31]

Answer: The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

Explanation:

Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

$\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M$

Hence, the molarity of sucrose solution is 0.146 M

Calculating the molality of solution:

To calculate the mass of solvent, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

$1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (sucrose) = 15 g

$M_{solute}$ = Molar mass of solute (sucrose) = 342.3 g/mol

$W_{solvent}$ = Mass of solvent = 339 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m$

Hence, the molality of sucrose solution is 0.129 m

5 0

4 years ago