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3 years ago

How is a gas's ability to fill a container different from that of a liquid or a solid?

1 answer:

8 0

Well a solid and a liquid is completely different from a gas. For a gas you need to have an air tight seal in order for the gas to stay in, or a special type of container so the gas doesn't decompose the container. And a liquid or a solid it's self explanatory. With some liquids you need special containers to hold the liquid so it doesn't decompose the container it's in.

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Fill in the coefficients that will balance the following reaction:

andrey2020 [161]

Answer:

CH4 + 2 O2 → CO2 + 2 H2O

Explanation:

There are one mole of O2 on the left side and on the right side there are three moles of O2. And to fix it you would need to make it two moles of O2 to have four molecules of O2 on the left side. Then you would make two moles of H2O to have a total of four moles of O2 on the right. Therefore, CH4 + 2 O2 → CO2 + 2 H2O is the answer.

7 0

3 years ago

Read 2 more answers

How can you chose values for x when making a table of values representing a real world situation?

Nitella [24]

X is usually representing time so years, days, etc

7 0

3 years ago

A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?

Rashid [163]

Answer:

Normality N = 0.2 N

Explanation:

Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.

Normality is represented by N.

Mathematically, we have :

$\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}$

Given that:

number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent

volume of solution (HCl) = 450 mL 450 × 10⁻³ L

$\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}$

Normality N = 0.2 N

7 0

3 years ago

How can we prevent land pollution?

Dominik [7]

Answer:

1. Make people aware about the concept of Reduce, Recycle and Reuse.

2. Reduce the use of pesticides and fertilizers in agricultural activities.

3. Avoid buying packages items as they will lead to garbage and end up in landfill site.

4. Ensure that you do not litter on the ground and do proper disposal of garbage.

5. Buy biodegradable products.

6. Do organic gardening and eat organic food that will be grown without the use of pesticides.

7. Create dumping ground away from residential areas.

I hope it helps u

7 0

2 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago