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Romashka [77]
3 years ago
5

In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il

lustrating how the PE and KE in a system changed as a skateboarder rides a halfpipe or a pendulum moves, why did the bar for the total energy remain constant?
A
This is because no energy is being created or destroyed in this system
B Because these models do not take into account the impact of friction and air resistance and are helping to solidify the concept of energy conservation and that the total mechanical energy remains constant in that model.
C Energy is converted from kinetic to potential and potential to kinetic, but the total amount of energy is conserved.
D
all answers given are correct
Chemistry
2 answers:
Lunna [17]3 years ago
7 0

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.        

Kazeer [188]3 years ago
7 0

Answer:..............

D

Explanation:

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If the ph of a solution is 7.0, what is the poh? <br> a. 14.0 <br> b. 0.0 <br> c. 7.0 <br> d. –7.0
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The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?
juin [17]

The formula for mole fraction is:

mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}    -(1)

The solubility of oxygen gas = 1.0 mmol/L  (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

Density = \frac{mass}{volume}

Mass = 1.0 g/ml\times 1000 mL = 1000 g

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = \frac{1000 g}{18 g/mol} = 55.55 mol

Substituting the values in formula (1):

mole fraction = \frac{0.001}{55.55+0.001}

mole fraction = 1.8\times 10^{-5}

Hence, the mole fraction is 1.8\times 10^{-5}.

7 0
3 years ago
The ph of 0.010 m aqueous aniline is 8.32. What is the percentage protonated?
LekaFEV [45]

Answer : The percentage aniline protonated is, 0.0209 %

Explanation :

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-8.32\\\\pOH=5.68

Now we have to calculate the hydroxide ion concentration.

pOH=-\log [OH^-]

5.68=-\log [OH^-]

[OH^-]=2.09\times 10^{-6}M

The equilibrium chemical reaction will be:

NH_3+H_2O\rightleftharpoons NH_4^++OH^-

From the reaction we conclude that,

Concentration of OH^- ion = Concentration of NH_4^+ ion = 2.09\times 10^{-6}M

Now we have to calculate the percentage aniline protonated.

\text{percentage aniline protonated}=\frac{2.09\times 10^{-6}M}{0.010M}\times 100

\text{percentage aniline protonated}=0.0209\%

Thus, the percentage aniline protonated is, 0.0209 %

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