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3 years ago

How would you prepare 100.0 ml of.400 m CaCl2 from a stock solution of 2.00 M CaCl2?

Chemistry

2 answers:

exis [7]3 years ago

7 0

Answer is: add 80.00 mL of water to 20.00 mL of stock solution.

V₁(CaCl₂) = ? mL; initial volume of calcium chloride solution.

c₁(CaCl₂) = 2.00 M; initial concentration.

V₂(CaCl₂) = 100.0 mL; final volume.

c₂(CaCl₂) = 0.400 M; final concentration.

Use c₁V₁ = c₂V₂.

2.00 M · V₁(CaCl₂) = 0.400 M · 100.0 mL.

V₁(CaCl₂) = (0.400 M · 100.0 mL) ÷ 2.00 M.

V₁(CaCl₂) = 20 ml.

V(H₂O) = 100 mL - 20 mL.

V(H₂O) = 80 mL.

docker41 [41]3 years ago

4 0

C₀=2 mol/l
c₁=0.400 mol/l
v₁=100.0 ml = 0.1 l

c₁v₁=c₀v₀

v₀=c₁v₁/c₀
v(H₂O)=v₁-v₀

v₀=0.1*0.400/2=0.02 l = 20 ml
v(H₂O)=100 - 20 = 80 ml

It is necessary to mix 20 ml of the feed solution and 80 ml of water.

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A balloon contains 2.0 L of air at 101.5 kPa. You squeeze the balloon to a volume of .75 L. What is the pressure of air on the i

Nataly_w [17]

You would use the formula for Boyle's Law:

(P1) (V1) = (P2) (V2)

(101.5) (2.0) = (P2?) (.75)

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3 0

2 years ago

Read 2 more answers

How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2

jeyben [28]

<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

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

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol$

For the given chemical reaction:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = $\frac{1}{1}\times 0.00286mol=0.00286mol$ of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

$0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g$

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

4 0

3 years ago

Hydrogen boils at 20k. what is the boiling point of hydrogen on the celsius scale

Dimas [21]

Celsius scale is related to kelvin scale by the following equation,

⁰C = K-273

°C = K-273

So as here temperature is given in kelvin, so it can be converted into celsius as follows:

So 20 K = 20K-273 °C

= -253 °C .

So, the 20 K temperature equals to -253 °C.

So , -253 °C is equals to 20 K or 20 K temperature equals to -253 °C.

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3 years ago

Determine the possible traits of the calves if:

White raven [17]

Answer:

1. All red calves i.e. RR

2. All roan calves i.e RW

3. 2 red calves (RR) and two roan calves (RW)

Explanation:

According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):

1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR

2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW

3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).