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Ronch [10]
3 years ago
11

"What is the potential energy relative to the water surface of a diver at the top of a 26 m

Chemistry
1 answer:
Len [333]3 years ago
4 0

Answer:

PE = 8918 J

Explanation:

Formula for potential energy here is;

PE = mgh

Where;

m is mass

g is acceleration due to gravity

h is height

We are given;

m = 35 kg

h = 26 m

Thus;

PE = 35 × 9.8 × 26

PE = 8918 J

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What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?
Ymorist [56]

The   molarity  of (HNO₃) that was used  if 2.00 L must  be used   to prepare 4.5 L  of a 0.25M HNO₃ solution is   0.563 M


 <u><em>calculation</em></u>

  This is calculated  usind  M₁V₁=M₂V₂  formula

where,

         M₁(  molarity ₁) = ?

         V₁( volume ₁) = 2.00 L

        M₁ (molarity ₂) = 0.25M

         V₂( volume₂) = 4.5 L

make M₁ the subject  of the formula by  diving both side of the formula  by V₁

   M₁  is therefore = M₂V₂/V₁

M₁ =[ (0.25 M  x 4.5 L) / 2.00 L ]  =0.563 M

5 0
3 years ago
Who uses chemicals ?
nordsb [41]
Is there answer choices ??
4 0
3 years ago
Read 2 more answers
Why is it important to conserve or save our energy resources?
earnstyle [38]

answer:

  • reducing energy use limits the number of carbon emissions in the environment
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explanation:

  • credits: online research
5 0
3 years ago
What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

<u>Calculation:</u>

The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

6 0
3 years ago
Which is the odd one brass,copper, aluminum,iton , sapphire​
zavuch27 [327]

Sapphire.

It  is a stone not a metal.

4 0
3 years ago
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