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3 years ago

9

The volume of gas occupied 240.0ml when the pressure is 1.20 atm what volume at constant pressure.will the gas occupies when the

pressure is decreased to 0.860 atm ?

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

334.89 ml

Explanation:

V2 = V1P1/P2

V2 = 240×1.2/0.86

V2 = 334.89 ml

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Which of the following solutions has the HIGHEST hydroxide-ion concentration?a) 0.1 M HClb) 0.1 M H2SOc) a buffer solution with

Alexus [3.1K]

Answer:

c) a buffer solution with pH= 5

Explanation:

The equation of hydrolisis of water is:

1x10⁻¹⁴ = [H⁺] [OH⁻]

That means if we know [H⁺] we can find hydroxide-ion concentration and vice versa.

a) HCl 0.1M = [H⁺] = 0.1M;

1x10⁻¹⁴ / [H⁺]

1x10⁻¹⁴/ 0.1M = [OH⁻]

1x10⁻¹³M = [OH⁻] hydroxide- ion concentration.

b) 0.1M H₂SO₄ = 0.2M [H⁺]

1x10⁻¹⁴/ 0.2M = [OH⁻]

5x10⁻¹⁴M = [OH⁻]

c) pH = 5.

[H⁺] = 1x10⁻⁵M

1x10⁻¹⁴/ 1x10⁻⁵M = [OH⁻]

1x10⁻⁹M = [OH⁻]

d) pOH = 12.

[OH⁻] = 1x10⁻¹²M

That means risht answer is:

<h3>c) a buffer solution with pH= 5</h3>

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4 years ago

Which kinds of bonds are present in a single water molecule?.

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The type of bonding present in water (H2O) is hydrogen bonding.

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A 50.0 mL sample of waste process water from a food processing plant weighing 50.1 g was placed in a properly prepared crucible

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Answer:

TS (total solid, mg/L) = 57,174 mg/L

VS (volatile solid, mg/L) = 24,088 mg/L

Explanation:

First step to solve this problem is to know data and questions:

Data:

Sample volume = 50 mL

Sample weight = 50.1 g

P1 (weight empty crucible) = 17.1234 g

P2 (weight after drying a 103º in oven) = 19.9821 g

P3 (weight after incinatrion 550º) = 18.7777 g

Questions:

TS = ?

VS = ?

Formula:

We need to use formulas to calculate total solid (TS) and volatile solid (VS), these are:

$TS =\frac{(P2-P1)*\frac{1,000 mg}{1g} }{Sample volumen (L)}$

$VS = \frac{(P2-P3)*\frac{1,000 mg}{1g} }{Sample volume (L)}$

We have to transform mL to L so we will divide mL by 1,000 the sample volume:

$Sample Volumen (L) = 50 mL * \frac{1L}{1,000 mL} = 0.05 L$

In the formula the value of 1,000 results by the convertion factor to transform grams to miligrams (we have to multiplie by 1,000)

Now we need to replace data on previous formulas and we will get TS and VS expressed in mg/L:

$TS = \frac{(19.9821 g-17.1234g)*\frac{1,000 mg}{1g} }{0.05L} = 57,174mg/L$

$VS = \frac{(19.9821g-18.7777g)*\frac{1,000mg}{1g} }{0.05L}=24,088 mg/L$

We divide TS and VS formula by sample volume because the exercise is asking us to express the results in mg/L.

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