Rt apart complete calculate [h+] for [oh−]=4.1×10−4 m. express your answer using two significant figures. [h+][ h + ] = 2.4×10−11 m previous answers correct the formula for the ion-product constant of water, kw, is [oh−][h+]= 1×10−14. when rearranged to solve for [h+], it becomes [h+]===kw[oh−]1.0×10−144.1×10−42.4× 10−11m part bpart complete calculate [h+] for [oh−]=8.5×10−9 m. express your answer using two significant figures. [h+][ h + ] = 1.2×10−6 m previous answers correct part cpart complete calculate [h+] for a solution in which [oh−] is 100 times greater than [h+]. express your answer using two significant figures. [h+][ h + ] = 1.0×10−8 m previous answers correct since the [oh−] is 100 times greater than the [h+], this expression, [oh−]=100[h+], can be substituted into the ion-product constant expression: kw==[h+]×100[h+]100[h+]2 rearranging the equation to solve for [h+]: [h+]===(kw100)1/2(1.0×10−14100)1/21 .0×10−8m part d indicate whether the solution is acidic, basic, or neutral.

A: [H⁺] = 2.4 × 10⁻¹¹ M

B: [H⁺] = 1.2 × 10⁻⁶ M

C: [H⁺] =  1.0 × 10⁻⁸ M

D: A and C are basic and B is acid.

Explanation:

Part A: Calculate [H⁺] for [OH⁻] = 4.1×10⁻⁴ M.

We know the ion-product of water Kw is:

Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]

Then,

Part B: Calculate [H⁺] for [OH⁻] = 8.5×10⁻⁹ M.

Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]

Part C: Calculate [H⁺] for a solution in which [OH⁻] is 100 times greater than [H⁺].

We know that [OH⁻] = 100 . [H⁺]. If we replace this in the ion-product of water:

Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]

1.0 × 10⁻¹⁴ = [H⁺]. 100. [H⁺]

1.0 × 10⁻¹⁴ = 100 . [H⁺]²

1.0 × 10⁻¹⁶ = [H⁺]²

[H⁺] = √1.0 × 10⁻¹⁶ = 1.0 × 10⁻⁸ M

Part D: Indicate whether the solution is acidic, basic, or neutral.

A solution is acid when [H⁺] > 10⁻⁷ M, basic when [H⁺] < 10⁻⁷ M and neutral when [H⁺] = 10⁻⁷M. So A and C are basic and B is acid.

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c, liquids evaporate, creating a gas