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3 years ago

PLEASE HELP THIS IS DUE IN AN HOUR!!!

Chemistry

1 answer:

Soloha48 [4]3 years ago

7 0

Explanation:

The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH. [H3O+] = 10-pH or [H3O+] = antilog (- pH) Example: What is the hydronium ion concentration in a solution that has a pH of 8.34? On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34).

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Help I’m stuck on this question :

Dimas [21]

Answer:

The first ionization energy is the energy it takes to remove an electron from a neutral atom.

hope it is helpful :)

8 0

3 years ago

What is the mass of volume= 5.4 mL density= 2.5 g/mL

ddd [48]

Answer:

13.5g

Explanation:

Mass is defined as the measure of the amount of matter in an object. Its unit is kg or g.

Mass can be calculated using the formula:

m= d × v

where,

d= density

m= mass

v= volume

m= 2.5×5.4

m= 13.5g

5 0

3 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

What the answer question

Serga [27]

Answer:

it's a trigonal bipyramidial

Explanation:

because NH3 have 3 hydrogen atoms

4 0

3 years ago

How much heat in kj is released by burning 9.5 grams of methane?

tatyana61 [14]

Methane is the compound CH4, and burning it uses the reaction:

CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.

Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.

n = 9.5/16.042 = 0.592195 mol

Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>

7 0

3 years ago