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Luden [163]
2 years ago
8

Help with this worksheet please (24pts)

Chemistry
1 answer:
Igoryamba2 years ago
7 0
  • Atoms that loss or gain electrons are called ions. There are two types of ions: cations and anions.
  • Here, 2+ represents that Calcium (Ca) should gain 2 more electrons so that its number equals to that of protons.
  • 3- represents that Nitrogen (N) should loss 3 electrons to equivalent with the number of protons.
  • Here, anions are S^2-, P^3-, Se^2-, Br^-.
  • While cations are Cr^3+, Ag^+, Li^+, Ba^2+.
  • The number of protons of Magnesium (Mg) = 12
  • The number of nuetrons of Mg = 12
  • The number of electrons of Mg = 12

Hope you could get an idea from here.

Doubt clarification - use comment section.

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ke prop to v^2


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A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under
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Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

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