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nlexa [21]
3 years ago
12

when you put a book on a table the table pushes on the book is that newton's 1st,2nd,or 3rd law of motion ? 

Chemistry
1 answer:
ohaa [14]3 years ago
8 0
The table pushes the book and the book pushes the table
It's 3rd law
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Calculate the density in g/ml of 2.0 L of gasoline that weighs 1.32 kg
sveta [45]

Answer: 0.66 g/mL

Explanation: The formula for density is d= m/v where m=mass and v=volume.

The mass was given in the problem, m= 1.32kg

The volume was also given in the problem, v= 2.0L

d= m/v ---> d= 1.32kg/2.0L ---> d=0.66kg/L

The problem calls for the answer to be written in g/mL, so we must convert the units. 1 kg contains 1000g and 1 L contains 1000 mL.

1000g/1000mL = 1 so the units change but the 0.66 does not. Therefore, your answer is 0.66g/mL (Two significant figures because 2.0 only has 2 significant figures).

4 0
2 years ago
Write a conversion factor that converts between moles of calcium ions in calcium phosphide, Ca3P2, and moles of Ca3P2.
cricket20 [7]

The conversion factor between moles of calcium ion and moles of Ca_3P_2 in  Ca_3P_2 would be 3 to 1.

<h3>Mole fraction</h3>

Ca_3P_2 if formed from 3 carbon atoms and 3 phosphate atoms according to the equation:

3Ca + 2P --- > Ca_3P_2

Thus, there are 3 moles of calcium in every 1 mole of  Ca_3P_2.

The conversion factor between moles of calcium and moles of  Ca_3P_2 will. therefore, be 3 to 1 or simply 3:1.

More on mole fractions can be found here: brainly.com/question/8076655

#SPJ1

6 0
3 years ago
How much of the sulfur burnt is normally oxidized to hexavalent state?
jeyben [28]

I feel that it is D but I am not 100% sure sadly.

7 0
3 years ago
What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?
coldgirl [10]

Answer:

pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,2

Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:

7,0 = 7,2 + log₁₀ \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

Ratio obtained is:

0,63 = \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H₂PO₄⁻   ⇄   HPO4²⁻ +        H⁺

0,1 M +x      0,063M -x  0,001M -x -<em>because the addition of H⁺ displaces the equilibrium to the left-</em>

Knowing the equation of equilibrium is:

K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}

Replacing:

6,20x10⁻⁸ = \frac{[0,063-x][0,001-x]}{[0,1+x]}

You will obtain:

x² -0,064 x + 6,29938x10⁻⁵ = 0

Thus:

x = 0,063 → No physical sense

x = 0,00099990

Thus, [H⁺] in equilibrium is:

0,001 M - 0,00099990 = 1x10⁻⁷

Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =

-log₁₀ [1x10⁻⁷] = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!

5 0
3 years ago
GIVING BRAINLIEST, FIVE STARS AND THANKS!!!!!!!!!!!!!!!!!!!!!!!!!!!
FinnZ [79.3K]

Answer:

<u>The advantages of nuclear energy:</u>

*produces low-cost energy

*it releases zero carbon emissions,

*there is a promising future for nuclear technology

*it has a high energy density

*it is reliable

hope this helps, have a great day! :D

6 0
3 years ago
Read 2 more answers
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