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SIZIF [17.4K]
3 years ago
10

A radiator contains 10 quarts of fluid, 30% of which is antifreeze. how much fluid should be drained and replaced with pure anti

freeze so that the new mixture is 40% antifreeze?
Chemistry
2 answers:
Svetllana [295]3 years ago
4 0

Answer:C on edge:)

Explanation:

amm18123 years ago
3 0
Let x be the volume of fluid removed and the volume of pure antifreeze that is added. The concentration of antifreeze in the fluid is 0.3, the concentration in pure antifreeze is 1 and that in the final solution is 0.4 The volume of the final solution is 10.
(10 - x)(0.3) + x = 10(0.4)
0.3 + 0.7x = 0.4
x = 1/7 quarts

The volume that should be drained is 1/7 quarts
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Naphthalene, commonly found in moth balls, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of naphthalene is 128 g
Vikki [24]

Answer:

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

Explanation:

Centesimal composition means that in 100 g of compound we have x g of the element. Therefore in 100 g of naphthalene we have:

93.7 g of C

6.3 g of H

Let's make a rule of three:

In 100 g of naphthalene we have 93.7 g of C  and 6.3 g of H

In 128 g of naphthalene we would have:

128 . 93.7 / 100 = 120 g of C

128.  6.3 / 100 = 8 g of H

We convert the mass to moles, by molar mass:

120 g . 1mol / 12 g = 10 moles C

8 g . 1mol/ 1g = 8 moles H

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

(The sub-index of each element is divided by the largest possible number)

6 0
3 years ago
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Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)
Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0
3 years ago
why it was important to use dishwashing gloves over nitrile gloves for distillation experiment and when IR plates are cleaned.
jeka57 [31]

Answer:

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8 0
2 years ago
How much energy is required to raise 10 grams of water by 10 degrees Celsius? (in Joules)
Georgia [21]

Answer:

420j

Explanation:

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=420j

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3 years ago
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Answer:

A

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