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Dmitriy789 [7]
2 years ago
10

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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Answer:

39.2 L at STP

Explanation:

Convert the grams to moles first by dividing 56.0 by the molar mass of O2 (32.0) then convert to volume by multiplying by 22.4.

= 39.2 L

5 0
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A chemist makes a solution of Ca(NO3)2 by dissolving 21.3 g Ca(NO3)2 in water to make 100.0 mL of solution. What is the concentr
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Answer:

[NO₃⁻ ] = 2.596 M

Explanation:

Ca(NO₃)₂ dissolves in water according to the following equation:

Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻

The moles of Ca(NO₃)₂ that dissolve is found as followed:

(21.3 g) / (164.10 g/mol) = 0.1298... mol

The number of NO₃⁻ ions are related to the above quantity by the molar ratio:

(0.1298 mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.2596...mol NO₃⁻

The concentration of the nitrate ions is then calculated:

[NO₃⁻ ] = (0.2596...mol) / (100.0ml) x (1000mL/L) = 2.596 M

6 0
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What occurs both inside where it's hot and on the earth where it's cool
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<span> "convective" transport</span>
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Calculate the millimoles of solute in 1.88 L of a 0.00713 M NaCN solution. millimoles:
konstantin123 [22]

Answer:

13.4 milimoles.

Explanation:

The following data were obtained from the question:

Volume = 1.88 L

Molarity = 0.00713 M

Millimoles of NaCN =?

Next, we shall determine the number of mole NaCN in the solution. This can be obtained as follow:

Molarity = mole /Volume

Volume = 1.88 L

Molarity = 0.00713 M

Mole of NaCN =?

Molarity = mole /Volume

0.00713 = moles of NaCN /1.88

Cross multiply

Mole of NaCN = 0.00713 × 1.88

Mole of NaCN = 0.0134 mole

Finally, we shall convert 0.0134 mole to Millimoles. This can be obtained as follow:

1 mole = 1000 millimoles

Therefore,

0.0134 mole = 0.0134 × 1000

0.0134 mole = 13.4 milimoles

Therefore, the millimoles of the solute, NaCN in the solution is 13.4 milimoles

7 0
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Answer: Option A

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Halogens(cl) always shows -1 oxidation state and IA group elements(Na)shows +1 oxidation state

4 0
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