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2 years ago

A sample of gas has a volume of 400 liters when its temperature is 20.C and its pressure

Chemistry

1 answer:

ikadub [295]2 years ago

5 0

Answer:

V₂ = 145.35 L

Explanation:

Given data:

Initial volume = 400 L

Initial pressure = 300 mmHg (300/760 =0.39 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂ /T₁ P₂

V₂ = 0.39 atm × 400 L × 273 K / 293 K × 1 atm

V₂ = 42588 atm.L.K /293 K.atm

V₂ = 145.35 L

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Transfer 2-3 drops of each solution onto the indicator paper. Be sure to

alexira [117]

Answer:

0.1 M HCl pH: 1-2

0.001 M HCl pH: 3-4

0.00001 M HCl pH: 5-6

Distilled Water pH: 7

0.00001 M NaOH pH: 8-9

0.001 M NaOH pH: 10-11

0.1 M NaOH pH: 12-13

Explanation:

4 0

3 years ago

Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s

Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, $2.67\times 10^2mol/L$

Explanation :

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = molar solubility of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 0.2 atm = 1.97×10⁻⁶ Pa

$k_H$ = Henry's law constant = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

$C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)$

$C_{O_2}=8.55\times 10^3g/L$

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = $\frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L$

Therefore, the molar concentration of oxygen is, $2.67\times 10^2mol/L$

8 0

3 years ago

Can someone me Help please

soldier1979 [14.2K]

Independent would be the amount of sugar given and the dependent would be the amount of cavities

Scientists have control groups so they can have evidence to show the difference between that one and the experimental ones.

And the independent variable is the one that changes because with out it changing you wouldn't get different results.

I could be wrong tho sorryyy! but i hope this helps

8 0

3 years ago

A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

balu736 [363]

Answer:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

4 0

3 years ago

Read 2 more answers

How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose

katen-ka-za [31]

Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

6 0

2 years ago