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.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j