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IrinaK [193]
2 years ago
15

A ball of mass 0.2 kg is dropped from a height of 10 m. How much mechanical energy does it have right before it hits the ground?

(Assume there is no air resistance.) Acceleration due to gravity is g 9.8 m/s2.
(PLEASE HELP NEED ANSWERS ASAP)

Chemistry
1 answer:
lana [24]2 years ago
3 0

Answer:

19.6 J  

Step-by-step explanation:

Before the ball is dropped, it has a <em>potential energy </em>

PE = mgh

PE = 0.2 × 10 × 9.8

PE = 19.6 J

Just before the ball hits the ground, the potential energy has been converted into kinetic (<em>mechanical</em>) energy.

KE = 19.6 J

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L
PtichkaEL [24]

Answer:

  9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

 P₁ = 98.7 Kpa

 T₂ = 20°C = 20 + 273.15 K = 293.15 K

  P₂ = 102.7 KPa

  V₂ = ?

2) Formula:

Used combined law of gases:

  PV / T = constant

  P₁V₁ / T₁ = P₂V₂ / T₂

3) Solution:

Solve the equation for V₂:

  V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ =  9.4 liter ← answer

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Explanation:

7 0
2 years ago
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prohojiy [21]

Answer:

HNO₃.

Explanation:

  • It is known that acids decrease the pH of the solution, while bases increase  the pH of the solution.

So, HF and HNO₃ decrease the pH of the solution as they produce H⁺ in the solution.

While, KOH and NH₃ increase the pH of the solution as they produce OH⁻ in the solution.

HNO₃ will decrease the pH of the solution greater than HF.

  • Because HNO₃ is strong acid that decomposes completely to produce H⁺ more than the same concentration of HF that is a weak acid which does not decomposed completely to produce H⁺.
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