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STatiana [176]
3 years ago
5

As the temperature of a liquid solvent increases, the amount of solute that can dissolve in it ________. decreases remains const

ant increases decreases by 1º celsius for every milliliter of solvent
Chemistry
1 answer:
Nookie1986 [14]3 years ago
6 0
Answer:
              
<span>As the temperature of a liquid solvent increases, the amount of solute that can dissolve in it <u>increases</u>.

Explanation:
                    The solubility of most solutes in a solvent increases with increase in temperature. This solubility is closely related to the heat of solution, (the heat evolved or absorbed when solute is dissolved in solvent). Hence, majority of solutes when dissolved in solvent absorbs heat and makes the overall heat of solution positive. Hence, in this case more heat provided will increase the rate of solubility.</span>
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Molarity is expressed as the number of moles of solute per volume of the solution. For example, we are given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH per 1 L volume of the solution. Acids and bases can be measured through the concentrations of H+ and OH- ions in units of molarity. Hope this helps.
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3 years ago
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Freon−12 (CF2Cl2), widely used as a refrigerant and aerosol propellant, is a dangerous air pollutant. In the troposphere, it tra
nevsk [136]

Answer:

141g of CCl₄

Explanation:

First, we have to write the balanced equation.

CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)

We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.

V = 14.9 dm³ = 14.9 L

T = 21°C + 273.15 = 294.15 K

P = 1.48 atm

R = 0.08206 atm.L/mol.K

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.48atm.14.9L}{0.08206\frac{atm.L}{mol.K}.294.15K }=0.914mol

We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.

0.914molCF_{2}Cl_{2}.\frac{1molCCl_{4}}{1molCF_{2}Cl_{2}} .\frac{154gCCl_{4}}{1molCCl_{4}} =141gCCl_{4}

7 0
3 years ago
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shutvik [7]

Answer:

look at the picture

Explanation:

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2 years ago
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5 0
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I hope this helps.  Let me know if anything is unclear.



7 0
3 years ago
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