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3 years ago

At room temperature, all samples of liquid water must have the same

Chemistry

2 answers:

Nikolay [14]3 years ago

6 0

The answer is 2. density

At room temperature, all samples of liquid water must have the same density.

Whatever the volume of different liquid water samples, they always have same density. As density of liquid water is always same that is 1 g cm⁻³. So different liquid water can have different volume, weight and mass but their density always remains same.

laiz [17]3 years ago

4 0

Density. water's density is always 1 g/ cm^3

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Yuliya22 [10]

Answer:

the name for NO is nitrogen monoxide

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2 years ago

What is the pOH of a 0.025 M [H+] solution? A.0.94 B.1.60 C.12.40 D.13.06

Greeley [361]

Answer:

option C= 12.40

Explanation:

Formula:

pH + pOH = 14

First of all we will calculate the pH.

pH = - log [H⁺]

pH = - log [0.025]

pH = - (-1.6)

pH = 1.6

Now we will put the values in formula,

pH + pOH = 14

pOH = 14-pH

pOH = 14 -1.6

pOH = 12.4

The pOH of solution is 12.4.

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2 years ago

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3 0

3 years ago

(Thermodynamics)

frutty [35]

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹

1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

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3 0

3 years ago

When copper metal is added to nitric acid, the following reaction takes place

zlopas [31]

Answer:

The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.

Explanation:

$Cu (s) + 4 HNO_3 (aq) \rightarrow Cu(NO_3)_2 (aq) + 2 H_2O (l) + 2 NO_2 (g)$

Moles of copper = $\frac{2.01 g}{63.55 g/mol}=0.03163 mol$

According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.

Then 0.03613 moles of copper will give:

$\frac{2}{1}\times 0.03163 mol=0.06326 mol$ of nitrogen dioxide gas

Moles of nitrogen dioxide gas = n = 0.06326 mol

Pressure of the gas = P

P = Total pressure - vapor pressure of water

P = 726 mmHg - 23.8 mmHg = 702.2 mmHg

P = 0.924 atm (1 atm = 760 mmHg)

Temperature of the gas = T = 25.0°C =298.15 K

Volume of the gas = V

$PV=nRT$

$V=\frac{0.06326 mol\times 0.0821 atm L/mol K\times 298.15 K}{0.924 atm}$

V = 1.68 L

The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.

3 0

3 years ago