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Why was 1990 an important year regarding air quality?

Valentin [98]

Why was 1990 an important year regarding air quality? Check all that apply.

2.Cost-effective ways to reduce pollution were emphasized.

4.Modifications and improvements were made to the Clean Air Act.

i just took the test this is 100% correct

3 0

3 years ago

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Difference between NaOH and HCI

morpeh [17]

Compare HCl, NaOH, and NaCl: HCl is a stronger acid than water. NaCl is a weaker base than NaOH. Strong acids react with strong bases to form weaker acids and bases. ... Compare NaOH, NH3, and H2O, and NH4Cl: NaOH is a stronger base than NH3.

7 0

3 years ago

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A scientist carries out a reaction twice. In Trial A, the scientist combines the reactants in a container. In Trial B, the scien

Paha777 [63]

It would be C because It will have a lower activation energy than Trial A.

7 0

3 years ago

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Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M A

olchik [2.2K]

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

<em>Where the concentrations [] are in equilibrium</em>

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

<em>Where the concentrations [] are the actual concentrations</em>

<em />

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

<em>Where X is defined as the reaction coordinate</em>

<em />

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

6 0

3 years ago

determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass

Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

$K_{f}$ or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

Δ $T_{f}$ = i × $K_{f}$ ×m

T⁰- $T_{s}$ = $i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water in kg}$

in which,

Δ $T_{f}$ = change in freezing point

Δ $T_{s}$ = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T $_{s}$ = 1 ×1.86°C/m × $\frac{30.7}{92.09g/mol * 0.376kg}$

$T_{s}$ = -1.65°C

Thus, the freezing point depression of a solution is -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

Learn more about freezing point depression :

brainly.com/question/26525184

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8 0

9 months ago