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2 years ago

HELP! ASAP!

Chemistry

1 answer:

harina [27]2 years ago

8 0

Given the model from the question,

The products are: N₂, H₂O and H₂
The reactants are: H₂ and NO
The limiting reactant is H₂
The balanced equation is: 3H₂ + 2NO —> N₂ + 2H₂O + H₂

<h3>Balanced equation </h3>

From the model given, we obtained the ffolowing

Red => Oxygen
Blue => Nitrogen
White => Hydrogen

Thus, we can write the balanced equation as follow:

3H₂ + 2NO —> N₂ + 2H₂O + H₂

From the balanced equation above,

Reactants: H₂ and NO
Product: N₂, H₂O and H₂

<h3>How to determine the limiting reactant</h3>

3H₂ + 2NO —> N₂ + 2H₂O + H₂

From the balanced equation above,

3 moles of H₂ reacted with 2 moles of NO.

Therefore,

5 moles of H₂ will react with = (5 × 2) / 3 = 3.33 moles of NO

From the calculation made above, we can see that only 3.33 moles of NO out of 4 moles given are required to react completely with 5 moles of H₂.

Thus, H₂ is the limiting reactant

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

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3 years ago

20.352 mL of chlorine under a pressure of 680. mm Hg are

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Answer:

0.01144L or 1.144x10^-2L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 20.352 mL

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P2 (final pressure) = 1210mmHg

V2 (final volume) =.?

Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:

P1V1 = P2V2

680 x 20.352 = 1210 x V2

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V2 = (680 x 20.352)/1210

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What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

<u>Step 1: </u>The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

<u>Step 2: </u>Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

<u>Step 3: </u>Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

<u>Step 4: </u>Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

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3 years ago

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

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