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Fudgin [204]
3 years ago
9

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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2 years ago
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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

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3 years ago
Kendra conducts an experiment. She mixes two chemicals, resulting in a chemical reaction. Based on Kendra's
kondaur [170]

Answer: It changed in identity and properties.

Explanation:

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3 years ago
What is the temperature of 1.2 moles of helium gas at 2.57 ATM if it occupies 15.5L of volume
agasfer [191]
We will assume helium to behave as an ideal gas and apply the ideal gas law:
PV = nRT
For pressure measured in atmospheres and volume measured in liters, the value of the molar gas constant is 0.082. Therefore:
T = PV / nR
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8 0
3 years ago
Give the n and l values and the number of orbitals for sublevel 5g.
Pepsi [2]

The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

There are total four quantum numbers:

1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4     ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1)  ( 2×4 + 1) = 9

Thus,  The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

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