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Levart [38]
3 years ago
13

Calculate hydrochloric acid (umol)in 200 ul of a 0.5173Msolution of acid?

Chemistry
1 answer:
strojnjashka [21]3 years ago
7 0

<u>Answer:</u> The moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Or,

\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}

We are given:

Molarity of solution = 0.5173 M

Volume of solution = 200\mu L

Putting values in above equation, we get:

0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol

Hence, the moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

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What is the net ionic equation for 2AgNO3 + 2NaOH = Ag2O + 2NaNO3 + H2O
soldi70 [24.7K]

Answer:

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

Explanation:

Step 1: RxN

2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O

Step 2: Define states of matter

2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)

Step 3: Total Ionic Equation

2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)

Step 4: Cancel out spectator ions

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

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3 years ago
Which best describes the future of wind power for providing electricity in the United States?
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A multivitamin tablet contains 40 milligrams of potassium. how many moles of potassium does each tablet contain?
katrin2010 [14]
Conversion of mole to grams
k in mole = 1 mole/ atomic mass
K in mole =1/ 39.0983 g/mole
                = 0.255765 g/mole
converting 40 grams of K
 K 40 grams x [ 1 mole/ 39.0983 grams] = 1.0230623 mole
There are 1.0230623 moles of K in 40 K of Potassium

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3 years ago
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Sodium metal reacts with water to produce hydrogen gas.
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Answer: I just took the test. The answer is D! (A single replacement reaction takes place because sodium is more reactive than hydrogen.)

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Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det
mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0
3 years ago
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