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slega [8]
3 years ago
11

6. How many moles of Cu(NO3)2 are in 25 ml of a 0.35 M solution of Cu(NO3)2? a) How many moles of NO3- are present in 25 ml of t

he above solution? b) What volume of this solution should be used to get 0.050 moles of Cu(NO3)2?

Chemistry
2 answers:
Pavlova-9 [17]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

Harman [31]3 years ago
4 0

Answer:

0.00875 moles Cu(NO3)2

a) 0.0175 moles NO3-

b) We should use 0.14L

Explanation:

Step 1: Data given

Volume = 25 mL = 0.025 L

Molarity of a Cu(NO3)2 solution = 0.35 M

Step 2: Calculate moles Cu(NO3)2

Moles Cu(NO3)2 = molarity * volume

Moles Cu(NO3)2 = 0.35 M * 0.025 L

Moles Cu(NO3)2 = 0.00875 moles

Step 3: Calculate moles NO3-

Cu(NO3)2 → Cu^2+ + 2NO3-

In 1 mol Cu(NO3)2 we have 2 moles NO3-

For 0.00875 moles Cu(NO3)2 we'll have 2*0.00875 = 0.0175 moles NO3-

Step 4: What volume of this solution should be used to get 0.050 moles of Cu(NO3)2?

Volume =  moles / molarity

Volume = 0.050 moles / 0.35 M

Volume = 0.14L

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How many milliliters of 0.125 M FeCl3 are needed to react with an excess of Na2S to produce 3.75 g of Fe2S3 if the percent yield
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Answer:

0.912 mL

Explanation:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

FeCl3 is the limiting reactant.

Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles

Hence actual yield of Iron III sulphide = 0.043 moles

Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield

Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide

From the reaction equation,

2moles of iron III chloride produced 1 mole of iron III sulphide

x moles of iron III chloride, will produce 0.057 of iron III sulphide

x= 2× 0.057= 0.114 moles of iron III chloride

But

Volume= number of moles/ concentration

Volume= 0.114/0.125

Volume= 0.912 mL

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What causes a spring to bounce up when it is pushed down and then released?
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(a) Pushing the spring down gives it stored mechanical energy that turns into motion

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Pushing on the spring causes the mechanical energy, of pushing on the spring, to be stored in the spring through potential elastic energy. Due to the elasticity of the spring, when the spring is released and resumes its initial shape the stored energy is released and can be used to do work such as motion.

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Every spring has an equilibrium position. Which statements describe a spring at its equilibrium position? Check all that apply.
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The statements describe a spring at its equilibrium position is the elastic potential energy is zero, the displacement of the spring is at a maximum and the net force acting on the spring is zero.

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In this case the only statements that match the definition of equilibrium position are:

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2 years ago
A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been adde
hram777 [196]

The pH of the solution after addition of 28 mL of NaOH is added to HCl is \boxed{{\text{0}}{\text{.85}}} .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

{\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left({\text{L}} \riht){\text{ of solution}}}}          

                             ......(1)        

                         

Rearrange equation (1) to calculate the moles of solute.

{\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}}\right)\left({{\text{Volume of solution}}}\right)       ......(2)

Substitute 0.50 M for the molarity of solution and 50 mL for the volume of solution in equation (2) to calculate the moles of HCl.

\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HCl}}&= \left({{\text{0}}{\text{.50 M}}}\right)\left( {{\text{50 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.02{\text{5 mol}}\\\end{aligned}

Substitute 0.50 M for the molarity of solution and 28 mL for the volume of solution in equation (2) to calculate the moles of NaOH.

\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{NaOH}}&=\left( {{\text{0}}{\text{.50 M}}} \right)\left( {{\text{28 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&= 0.014{\text{ mol}}\\\end{aligned}

The reaction between HCl and NaOH occurs as follows:

{\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}

The balanced chemical reaction indicates that one mole of NaOH reacts with one mole of HCl. So the amount of remaining HCl can be calculated as follows:

\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&= {\text{0}}{\text{.011 mol}} \\\end{aligned}

The volume after the addition of NaOH can be calculated as follows:

\begin{aligned}{\text{Volume of solution}} &= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}

Substitute 0.011 mol for the amount of solute and 78 mL for the volume of solution in equation (1) to calculate the molarity of new HCl solution.

\begin{aligned}{\text{Molarity of new HCl solution}}&= \left({{\text{0}}{\text{.011 mol}}} \right)\left( {\frac{1}{{{\text{78 mL}}}}}\right)\left( {\frac{{{\text{1 mL}}}}{{{{10}^{ - 3}}\;{\text{L}}}}} \right)\\&= 0.1410{\text{2 M}}\\&\approx {\text{0}}{\text{.141 M}}\\\end{aligned}

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

{\text{pH}}=- {\text{log}}\left[ {{{\text{H}}^ + }}\right]     ......(3)

Here,

\left[{{{\text{H}}^ + }}\right] is hydrogen ion concentration.

HCl is a strong acid so it dissociates completely. So the concentration of   also becomes 0.141 M.

Substitute 0.141 M for \left[{{{\text{H}}^ + }}\right] in equation (3).

\begin{aligned}{\text{pH}}&= - {\text{log}}\left({0.141} \right)\\&=0.85\\\end{aligned}

So the pH of the solution is 0.85.

Learn more:

1. Which indicator is best for titration between HI and  ? brainly.com/question/9236274

2. Why is bromophenol blue used as an indicator for antacid titration? brainly.com/question/9187859

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acid-base titrations

Keywords: molarity, pH, HCl, NaOH, 0.85, 0.141 M, moles of HCl, moles of NaOH, 50 mL, 0.50 M, 28 mL, 0.025 mol, 0.014 mol, 0.011 mol, 78 mL.

4 0
3 years ago
Read 2 more answers
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