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adell [148]
3 years ago
14

Chlorine + lithium iodide →

Chemistry
1 answer:
tino4ka555 [31]3 years ago
5 0

Answer:

Cl_{2} (g)+2Lil(aq) ---> 2LiCl(aq)+I_{2}

Explanation:

---> lithium chloride + iodine

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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL
d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

pH = 0.70

Putting values in above equation, we get:

0.70=-\log[H^+]

[H^+]=10^{-0.70}=0.199M

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated nitric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

We are given:

M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL

Hence, the volume of concentrated solution required is 9.95 mL

6 0
2 years ago
Electrical energy can produce many other types of energy.
Tomtit [17]
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7 0
2 years ago
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If a person is sterile, he or she is:
Tresset [83]

Answer:

1

Explanation:

Sterile means sperm or eggs are not able to Technichaly Create or produce.

7 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
For the chemical reaction 2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq) write the net ionic equation, including the phases.
Crazy boy [7]

Answer:

2H⁺(aq) + 2OH⁻(aq)  --> 2H2O(l)

Explanation:

2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)

We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.

The ionic equation is given as;

2H⁺(aq)  +  2Br⁻(aq)  + Ba²⁺(aq) + 2OH⁻(aq)   --> 2H2O(l)  +  Ba²⁺(aq)  + 2Br⁻(aq)

Upon eliminating the spectator ions; The net equation is given as;

2H⁺(aq) + 2OH⁻(aq)  --> 2H2O(l)

8 0
3 years ago
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