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castortr0y [4]
3 years ago
5

What happens if you were to swallow a small amount if wet nail polish? Answer truthfully in your own words.

Chemistry
1 answer:
Scilla [17]3 years ago
5 0

Answer:

I'm pretty sure you'll be fine

Explanation:

I've swallowed more than a small amount and I'm ok sort of

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1.42 g H2 is allowed to react with 10.4 g N2 , producing 2.14 g NH3 . Part A What is the theoretical yield in grams for this rea
Bad White [126]

Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 H₂ + N₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • H₂: 3 moles
  • N₂: 1 mole
  • NH₃: 2 moles

The molar mass of the compounds is:

  • H₂: 2 g/mole
  • N₂: 28 g/mole
  • NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • H₂: 3 moles ×2 g/mole= 6 grams
  • N₂: 1 mole ×28 g/mole= 28 grams
  • NH₃: 2 moles ×17 g/mole= 34 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?

mass of H_{2} =\frac{10.4 grams of N_{2}x 6 grams of H_{2} }{28 grams of N_{2}}

<u><em>mass of H₂= 2.2286 grams</em></u>

But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.

<h3>Definition of theorical yield</h3>

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Theoretical yield in this case</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?

mass of NH_{3} =\frac{1.42 grams of H_{2} x 34 grams of NH_{3}}{6grams of H_{2} }

<u><em>mass of NH₃= 8.0467 grams</em></u>

Then, the theorical yield for the reaction is 8.0467 grams of NH₃.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

7 0
1 year ago
Match the vocabulary word with the defination.
OLga [1]

Answer:

1. d

2. b

3. d

4. e

5. a

explanation:

there's nothing else to explain

3 0
3 years ago
Read 2 more answers
PLEASE HELP!! A gas has a volume of 7.9 L at a temperature of 222. K and pressure of 877 mmHg. What is the new temperature (in C
Natali5045456 [20]

Answer:

82.97 K

Explanation:

Applying,

PV/T = P'V'/T'................ Equation 1

Where P = initial pressure, T = Initial temperature, V = Initial Volume, P' = Final pressure, V' = Final Volume, T' = Final Temperature.

Make T' the subject of formula in equation 1

T' = P'V'T/PV................ Equation 2

From the question,

Given: P = 877 mmHg = (877×0.001316) atm = 1.154 atm, T = 222.2 K, V = 7.9 L, P' = 0.327 atm, V' = 10.41 L

Substitute these values into equation 2

T' = (0.327×10.41×222.2)/(1.154×7.9)

T' = 82.97 K

Hence the new temperature is 82.97 K

7 0
3 years ago
The atoms of some elements can be made
natima [27]

Answer:

I think 4 I got it right on edg but i duno if its the same

Explanation:

8 0
3 years ago
Calculate the volume occupied<br>at s.t.p by 6.89<br>gas [H = 10, N = 14​
Nadya [2.5K]

Answer:

9.07 L

Explanation:

<em>Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0​].</em>

Step 1: Given and required data

  • Mass of NH₃ (m): 6.89 g
  • Molar mass of NH₃ (M): 17.0 g/mol

Step 2: Calculate the moles (n) of NH₃

We will use the following epxression.

n = m / M

n = 6.89 g / (17.0 g/mol) = 0.405 mol

Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP

At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).

0.405 mol × 22.4 L/1 mol = 9.07 L

8 0
3 years ago
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