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2 years ago

What is the pOH of a solution with a pH of 8? 14 6 8 0

Chemistry

1 answer:

OverLord2011 [107]2 years ago

4 0

Answer:

Explanation:

∵ pH + pOH = 14.

∴ pOH = 14 - pH = 14 - 8 = 6.

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Which number label represents the nucleus? 6,5,3,2

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Explanation:

Nucleus is the center or "CEO" of the cell, that holds the cell's DNA and controls the cells' function.

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2 years ago

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Even though we are well into the course Dumbledore notices some of the students are still having trouble with significant figure

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Answer:

B) 2

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

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The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

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(3.478-2.31) = 1.168 ≅ 1.17 (Rounded to least decimal digit)

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3 years ago

What is a net ionic equation?

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Answer:

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3 years ago

A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h

dexar [7]

Answer:

5.758 is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d = $998.2 kg/m^3$

1 kg = 1000 g

$1 m^3=10^6 cm^3$

$998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3$

Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3$

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' = 9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

$v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3$

Density of metal = d'

Volume of metal = v' = $\frac{M'}{d'}$

Difference in volume will give volume of metal ingot.

v' = v - V

$v'=19.73 cm^3-18.08 cm^3=$

$v'=1.65 cm^3$

Since volume cannot be in negative .

Density of the metal =d'

= $d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3$

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