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Oduvanchick [21]
3 years ago
5

In an experiment designed to determine the concentration of Cu 2 ions in an unknown solution, you need to prepare 100 mL of 0.10

M CuSO4 as your stock solution. How many grams of CuSO4(s) should you use
Chemistry
1 answer:
Alexxandr [17]3 years ago
8 0

Answer:

1.6 grams

Explanation:

We need to prepare 100 mL (0.100 L) of a 0.10 M CuSO₄ solution. The required moles of CuSO₄ are:

0.100 L × 0.10 mol/L = 0.010 mol

The molar mass of CuSO₄ is 159.61 g/mol. The mass corresponding to 0.010 moles is:

0.010 mol × (159.61 g/mol) = 1.6 g

We should use 1.6 grams of CuSO₄.

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An equilibrium mixture contains 0.750 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the
sasho [114]

1.302  moles of carbon dioxide would have to be added

<h3>Further explanation</h3>

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}

While the equilibrium constant is based on partial pressure

\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}

The value of Kp and Kc can be linked to the formula '

\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}

R = gas constant = 0.0821 L.atm / mol.K

=n = number of product coefficients-number of reactant coefficients

An equilibrium mixture: 0.750 moles of CO2 and H2, and 0.200 moles of CO and H2O

  • We determine Kc (constant concentration)

\displaystyle Kc=\frac{0.75.0.75}{0.2.0.2}\\\\Kc=14.1

  • the amount of carbon monoxide to 0.300 mol

Reaction :

                CO +H₂O ⇔ CO₂ + H₂

initially      0.2   0.2       0.75+x  0.75

reaction    0.1    0.1        0.1         0.1

product     0.3   0.3       0.65+x    0.65

\displaystyle Kc=\frac{(0.650+x)(0.65)}{0.3.0.3}\\\\14.1(0.09)=0.4225+0.65x\\\\1.269-0.4225=0.65x\\\\0.8465=0.65x\\\\x=1.302

<h3>Learn more</h3>

an equilibrium constant brainly.com/question/9173805

brainly.com/question/1109930

Calculate the value of the equilibrium constant, Kc

brainly.com/question/3612827

Concentration of hi at equilibrium

brainly.com/question/8962129

Keywords: constant equilibrium, Kc, concentration, product, reactant, reaction coefficient

3 0
4 years ago
Read 2 more answers
Please help me! I put the max brainly points!
Diano4ka-milaya [45]

Answer:

579 mL, .96 kPa, 34.21 c

Explanation:

8 0
3 years ago
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HELP! What are 5 changes for testing out a fidget spinner no websites
Sedbober [7]

Answer:

Explanation:

Can you explain a bit further? once u do, ill edit my answer and tell you :) But to use a fidget spinner, u have to put ur thumb at the bottom and index finger at the top, and spin it with you other fingers if thats what you meant

6 0
3 years ago
A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.
Alenkinab [10]

Answer:

\Delta V = 234.736\,mL

Explanation:

The quantity of moles of ethanol in the solution is:

n_{C_{2}H_{5}OH} = \left(\frac{597\,mL}{1000\,mL} \right)\cdot \left(8.35\,\frac{mol}{L} \right)

n_{C_{2}H_{5}OH} = 4.985\,mol

The mass and volume of ethanol in the solution are, respectively:

m_{C_{2}H_{5}OH} = (4.985\,mol)\cdot \left(46.07\,\frac{g}{mol} \right)

m_{C_{2}H_{5}OH} = 229.658\,g

V_{C_{2}H_{5}OH} = \frac{229.658\,g}{0.7893\,\frac{g}{mL} }

V_{C_{2}H_{5}OH} = 290.964\,mL

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

\Delta V = (525\,mL+597\,mL) - (597\,mL + 290.964\,mL)

\Delta V = 234.736\,mL

8 0
3 years ago
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BabaBlast [244]
Yes you are correct good job!
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3 years ago
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