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madreJ [45]
3 years ago
11

I NEED HELP PLEASE, THANKS SO MUCH! :)

Chemistry
1 answer:
Anon25 [30]3 years ago
7 0

Answer:

Here's what I get  

Explanation:

The molecules of a substance attract each other.

They are also in constant motion.

Their kinetic energy increases with the temperature.

1. Solids

If the temperature is low enough, the molecules will have little kinetic energy.

They will not be able to escape the attractions of their neighbours, so they will be fixed in place in a crystalline array.

All they can do is vibrate about their position in the array.

The substance will be a solid.

2. Liquids

If the temperature is higher, the molecules will have enough kinetic energy to slide past each other, but not enough to escape the attractions of nearby molecules.

The substance will be a liquid.

3. Gases

If the temperature is high enough, the molecules will have enough kinetic energy to escape from the surface and fly off into space.

The molecules will on average so far apart that they won't feel the attractions of others.

When they do collide, they will have energy to bounce off each other rather than stick together.

The substance will be a gas.

 

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the
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Answer:

Choice A: approximately 0.12\; \rm M.

Explanation:

Note that the unit of concentration, \rm M, typically refers to moles per liter (that is: 1\; \rm M = 1\; \rm mol\cdot L^{-1}.)

On the other hand, the volume of the two solutions in this question are apparently given in \rm cm^3, which is the same as \rm mL (that is: 1\; \rm cm^{3} = 1\; \rm mL.) Convert the unit of volume to liters:

  • V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L.
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Calculate the number of moles of \rm H_2SO_4 formula units in that 0.02\; \rm L of the 0.09\; \rm M solution:

\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}.

Note that \rm H_2SO_4 (sulfuric acid) is a diprotic acid. When one mole of \rm H_2SO_4 completely dissolves in water, two moles of \rm H^{+} ions will be released.

On the other hand, \rm NaOH (sodium hydroxide) is a monoprotic base. When one mole of \rm NaOH formula units completely dissolve in water, only one mole of \rm OH^{-} ions will be released.

\rm H^{+} ions and \rm OH^{-} ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid \rm H_2SO_4 dissolves in water completely, it will take two moles of \rm OH^{-} to neutralize that two moles of \rm H^{+} produced. On the other hand, two moles formula units of the monoprotic base \rm NaOH will be required to produce that two moles of \rm OH^{-}. Therefore, \rm NaOH and \rm H_2SO_4 formula units would neutralize each other at a two-to-one ratio.

\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O.

\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2.

Previous calculations show that 0.0018\; \rm mol of \rm H_2SO_4 was produced. Calculate the number of moles of \rm NaOH formula units required to neutralize that

\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}.

Calculate the concentration of a 0.03\; \rm L solution that contains exactly 0.0036\; \rm mol of \rm NaOH formula units:

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