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AveGali [126]
3 years ago
13

How do you balance this equation? Show your work. Cu+HNO3=Cu(NO3)2+NO+H2O

Chemistry
1 answer:
GenaCL600 [577]3 years ago
7 0

3 Cu + 8 HNO3 = 3 Cu(NO3)2 + 2 NO + 4 H2O i dont know how to explain it but this is the balanced equation basically all the atoms of every element has to be balanced

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Can someone answer theses problems? Only do the Even number problems,Im in a rush so please help me.
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The answer to the questions. All 15 of them.

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4 years ago
Nucleotides have a phosphate group attached at the ___ carbon atom of the sugar.
Tanzania [10]

Answer:

Nucleotides have a phosphate group attached at the 5' carbon atom of the sugar.

Explanation:

Phosphate group is attached to the sugar molecule in the place of -OH.

6 0
3 years ago
The ka of phosphoric acid, h3po4, is 7.6  10–3 at 25 °c. for the reaction h3po4(aq) h2po4 – (aq) + h+ (aq) ∆h° = –14.2 kj/mol.
blsea [12.9K]
<span>Use the van't Hoff equation: ln ( K2 K1 ) = Δ Hº R ( 1 T1 ⒠1 T2 ) ln ( K2 7.6*10^-3 ) = -14,200 J 8.314 ( 1 298 ⒠1 333 ) ln ( K2 7.6*10^-3 ) = ⒠1708 ( 0.00035 ) ln ( K2 0.0076 ) = ⒠0.598 Apply log rule a = log b b a -0.598 = ln ( e ⒠0.598 ) = ln ( 1 e 0.598 ) Multiply both sides with e^0.598 K 2 e 0.598 = 0.0076 K e 0.598 e 0.598 = 0.0076 e 0.598 K 2 = 0.0076 e 0.598 = 4.2 ⋅ 10 ⒠3 K2 = 4.2 ⋅ 10 ⒠3</span>
7 0
4 years ago
A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0
3 years ago
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kenny6666 [7]

1333 KM is the sum,

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Read 2 more answers
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