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Semmy [17]
3 years ago
14

A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25

M solution could be made from the stock solution
Chemistry
1 answer:
mylen [45]3 years ago
5 0
<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

  • Concentration of stock solution as 1.00 M
  • Volume of the stock solution as 50 mL
  • Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

  • The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
  • Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

     = (1.00 M × 50 mL) ÷ 0.25 M

     = 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

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Naddika [18.5K]

Answer:

Percentage of carbon:

{ \tt{ =  \frac{24}{30}  \times 100\%}} \\  = 80\%

Percentage of hydrogen:

{ \tt{ =  \frac{6}{30}  \times 100\%} } \\  = 20\%

8 0
3 years ago
If the initial volume is 55ml and the final volume after some rocks of sandstone are added is 65ml what is the volume of the chi
AfilCa [17]

Answer:

Volume = 10ml

Density = 1/5 g/ml or 0.20g/ml

Explanation:

The rocks are 10ml since the initial volume went up by 10.

Since density = mass/volume, you divide 2 by 10.

D = 2/10

D = 1/5 g/ml or 0.20g/ml

(Unit is g/ml aka grams/millileter)

7 0
2 years ago
Plz help I’m pretty confused on this question
sergeinik [125]

Answer:

Bohr model A

Explanation:

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7 0
2 years ago
Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0
3 years ago
Which of the following elements will lose electrons to form an
photoshop1234 [79]
B and c...will lose electron(s) in forming an Ion.

P is an Anion
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d. Se is an Anion.
4 0
2 years ago
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