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Keith_Richards [23]

2 years ago

7

0.15 gm of metallic oxide was dissolved in 100 ml of 0.1 N H2SO4 and 25.8 ml of 0.095N NaOH were used to neutralise the remainin

g H2SO4.Calculate the equivalent weight of metallic oxide and metal.
Ans: metallic oxide=19.87
metal=11.87

1 answer:

Mila [183]2 years ago

7 0

Answer:

Explanation:

25.8 ml of .095 N NaOH is needed to neutralise the remaining acid

equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .

acid remaining = .002451 gm equivalent .

acid initially taken = 100 ml of .1 N / 1000 = . 01 gm equivalent

acid reacted with metal = .01 -.002451 = .007549 gm equivalent

This must have reacted with same gram equivalent of metal oxide

.007549 gm equivalent = .15 gm of metal oxide

1 gm equivalent = 19.87 gm

equivalent weight of metal = 19.87 - equivalent weight of oxygen

= 19.87 - 8 = 11.87 .

1

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. Divide 94.20 g by 3.167 22 mL.

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The density of o2 gas at 16 degrees Celsius and 1.27atm is?

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Answer:

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An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

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So, you can get:

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Then:

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Replacing:

$density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol} }{0.0821\frac{atm*L}{mol*K} *289 K}$

Solving:

density= 1.71 $\frac{g}{L}$

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1 year ago

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Putting values in above equation, we get:

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1 year ago

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2 years ago

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