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ahrayia [7]
3 years ago
6

How much water should be added to 70ml of 60 percent acid solution to dilute it to a 50 percent acid solution?

Chemistry
1 answer:
mestny [16]3 years ago
7 0

Answer:

14 ml of water

Explanation:

To find the volume you need to dilute the concentration of a solution, you should use the formula C1 x V1 = C2 x V2 in which:

C1 = initial concentration ( in this case 60 %)

V1 = initial volume ( in this case 70 ml)

C2 = Final concentration ( you want to dilute until 50 %)

V2 = final volume ( the variable you want to search)

So you need to:

1.- Isolate the variable you want to find: V2 = (C1 x V1) / C2

2.- Substitute data: V2 = (60% x 70 ml) /50 %

3.- You do the math, in this case is 84 ml.

4.- Remember that you have a initial volume of 70 ml, so the difference (84 ml - 70 ml = 14 ml) is the volume you need to add to dilute your solution.

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At 258C, Kp 5 2.9 3 1023 for the reactionNH4OCONH21s2m 2NH31 g21 CO21 g2In an experiment carried out at 258C, a certain amount o
iragen [17]

Answer:

0.27 atm

Explanation:

<em>At 25ºC, Kp = 2.9 x 10⁻³ for the reaction NH₄OCONH₂(s) ⇌ 2 NH₃(g) + CO₂(g). In an experiment carried out at 25ºC, a certain amount of NH₄OCONH₂ is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.</em>

Step 1: Make an ICE chart

Solid and liquids are ignored in ICE charts.

       NH₄OCONH₂(s) ⇌ 2 NH₃(g) + CO₂(g)

I                                            0             0

C                                        +2x           +x

E                                          2x             x

Step 2: Write the pressure equilibrium constant expression (Kp)

Kp = [NH₃]² × [CO₂]

Kp = (2x)² × x

2.9 × 10⁻³ = 4 x³

x = 0.090 atm

Step 3: Calculate the pressures at equilbrium

pNH₃ = 2x = 2(0.090 atm) = 0.18 atm

pCO₂ = x = 0.090 atm

The total pressure is:

P = 0.18 atm + 0.090 atm = 0.27 atm

4 0
2 years ago
Suppose a substance has a heat of fusion equal to 45 calg and a specific heat of 0.75
Varvara68 [4.7K]

Answer:

The substance will be in liquid state at a temperature of 97.3 °C

Note: The question is incomplete. The complete question is given below :

Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)

Explanation:

1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT

Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C

H = 50 × 0.75 × 3 = 112.5 calories

b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g

H = 50 × 45 = 2250 cal

c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.

Amount of energy left = 5000 - 2362.5 = 2637.5 cal

The remaining energy is used to heat the liquid

H = mcΔT

Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change

2637.5 = 50 × 0.75 x ΔT

ΔT = 2637.5 / ( 50*0.75)

ΔT = 70.3 °C

Final temperature of sample = (70.3 + 27) °C = 97.3 °C

The substance will be in liquid state at a temperature of 97.3 °C

5 0
3 years ago
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Lesechka [4]
Ca ionises into Ca^2+. Ca^2+ will be attracted to O^2- ions in the water, since opposite charges attract. (Hydrogen in water forms H^+)
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How can we prevent harmful algae blooms?

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Explanation:

I hope this helps

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