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Yuki888 [10]
4 years ago
15

Forrest spills some acid on the lab floor. He notifies his teacher and begins to look for baking soda to neturalize the acid on

the floor. Who should be notified next ?
Chemistry
2 answers:
Romashka [77]4 years ago
8 0

Explanation:

Since, acid is spilled on the lab floor so, first of all the teacher must be notified about it. As teacher can better tell you what should be done in such a case.

At the same time, rest of the students present in the lab must be notified about it so that no one can accidentally step onto the acid otherwise it might cause a serious injury to anyone.

Thus, we can conclude that rest of the students or classmates should be notified next in the given situation.

gregori [183]4 years ago
7 0
His classmates should be notified next
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How many acetone molecules are in a bottle of acetone with a volume of 445 mL ? (density of acetone = 0.788 g/cm3).
Mkey [24]
<h2>Hello!</h2>

The answer is: 3.63x10^{24}particles

<h2>Why?</h2>

First, we need to calculate the molecular weight of the acetone (CH3COCH3)

So,

C=12.011g/mol\\O=15.99g/mol\\H=1.008g/mol

Then,

CH3COCH3=12.011+(1.008)*3+12.011+15.999+12.011+(1.008)*3=58.08g/mol

Second, we need to calculate the mass of the acetone in the bottle

We need to remember that 1cm^{3}=1mL

So,

445cm^{3}*\frac{0.788g}{cm^{3}}=350.66g

Third, we need to calculate the number of moles:

350.66g*\frac{1mol}{58.08g}=6.03moles

Fourth, we need to calculate the number of atoms:

Remember, 1 mol=6.022x10^{23}particles

Therefore,

6.03moles*\frac{6.022x10^{23}particles }{1mol}=3.63x10^{24}particles

Have a nice day!

4 0
4 years ago
Read 2 more answers
A gas mixture consists of 4 kg of O2, 5 kg of N2, and 7 kg of CO2. Determine (a) the mass fraction of each component, (b) the mo
kondor19780726 [428]

Answer:

See explanation.

Explanation:

Hello

(a) In this case, one uses the following formulas, which allow to compute the mass fraction of each component:\% O_2=\frac{m_{O_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{4kg}{4kg+5kg+7kg}*100\%=25\%O_2\\\% N_2=\frac{m_{N_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{5kg}{4kg+5kg+7kg}*100\%=31.25\%N_2\\\% CO_2=\frac{m_{CO_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{7kg}{4kg+5kg+7kg}*100\%=43.75\%CO_2

(b) For the mole fractions, it is necessary to find all the components' moles by using their molar mass as shown below:

n_{O_2}=4kgO_2*\frac{1kmolO_2}{32kgO_2} =0.125kmolO_2\\n_{N_2}=5kgN_2*\frac{1kmolN_2}{28kgN_2} =0.179kmolN_2\\n_{CO_2}=7kgCO_2*\frac{1kmolCO_2}{44kgCO_2} =0.159kmolCO_2

Now, the mole fractions:

x_{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.125kmolO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=27\%O_2\\x_{N_2}=\frac{n_{N_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.179kmolN_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=38.7\%N_2\\ x_{CO_2}=\frac{n_{CO_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.159kmolCO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=34.3\%CO_2

(c) Finally the average molar mass is computed considering the molar fractions and each component's molar mass:

M_{average}=0.27*32g/mol+0.387*28g/mol+0.343*44g/mol=34.57g/mol

And the gas constant:

Rg=0.082\frac{atm*L}{mol*K}*\frac{1mol}{34.57g}=0.00237\frac{atm*L}{g*K}

Best regards.

8 0
4 years ago
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