3 years ago

W

Chemistry

1 answer:

natali 33 [55]3 years ago

8 0

Answer:

I would say is B but I do t really know

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What will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?

Marina86 [1]

Answer:

THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L

Explanation:

Using the ideal gas equation

PV = nRT

P = 1.75 atm

n = 2.5 moles

T = 475 K

R = 0.082 L atm/mol K

V = unknown

Substituting the variables into the equation we have:

V = nRT / P

V = 2.5 * 0.082 * 475 / 1.75

V = 97.375 / 1.75

V = 55.64 L

The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L

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Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s)

klemol [59]

Answer:

a) 2.01 g

Explanation:

Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

First we convert 0.0302 mol AgNO₃ to Na₂CO₃ moles, in order to calculate how many Na₂CO₃ moles reacted:

0.0302 mol AgNO₃ * $\frac{1molNa_2CO_3}{2molAgNO_3}$ = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

Finally we convert Na₂CO₃ moles into grams, using its molar mass:

0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).

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Explanation:

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