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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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what is the proper scientific notation for 0.000476? a.0.476 × 10-3 b.4.76 × 10-4 c.47.6 × 10-5 d.476 × 10-6
-BARSIC- [3]
B. 4.75 × 10-4 is the answer
6 0
3 years ago
Number of grams of hcl that can react with 0.500 grams of Al(OH)3
vlabodo [156]

Answer:

0.7g of HCl

Explanation:

First, let us write a balanced equation for the reaction between HCl and Al(OH)3.

This is illustrated below:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:

Molar Mass of Al(OH)3 = 27 + 3(16+1)

= 27 + 3(17) = 27 + 51 = 78g/mol.

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g

Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

From the equation above,

78g of Al(OH)3 reacted with 109.5g of HCl.

Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl

7 0
3 years ago
Indicate the number of significant figures in the measured number 0.0020 <br> A. 4<br> B. 5<br> C. 2
Sergio [31]

<u>Answer:</u>

The correct answer option is C. 2.

<u>Explanation:</u>

We are given the number '0.0020' and we are to indicate the number of significant figures in the given measured number.

According to the rules of significant figures, numbers that are non-zero, zeros between any two significant numbers and the ending zeros in the decimal position are categorized as significant figures.

Since there is one non-zero number and one ending zero in the decimal position, therefore 0.0020 has 2 significant figures.

6 0
3 years ago
A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a
trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 +  2 HCl  -->  MnCl2  +  H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl)  X (0.0020 mole HCl / 1000ml) X (4.86 ml)    

= 4.86 x 10^-3 mole  

this is now dissolved in (70 + 4.86)  =  74.86 ml or 0.07486 L

thus [Mn(OH)2]  =  4.86 x 10^-3 mole / 0.07486 L  =  0.064921 M

Ksp =  [Mn2+][OH-]^2  =  4x^3  =  4(0.064921)^3  = 10.945 x 10^-4

6 0
3 years ago
A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the vo
kozerog [31]

Answer:

0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

P1 = 717 mmHg

T1 = 315 K

V2 = ?

P2 = 463 mmHg

T2 = 777 K

The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.

(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)

0.346 = (463*V2) / (777)

0.346*777 = (463*V2) / (777)*777

268.842 = 463*V2

268.842/463 = (463*V2)/463

V2 = 0.581

Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.

6 0
3 years ago
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