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zmey [24]
3 years ago
13

A/an bonds allows metals to conduct electricity

Chemistry
1 answer:
dalvyx [7]3 years ago
5 0
If I understand you correctly, you should fill the gap and decide which kind og bonds <span> allows metals to conduct electricity. If so, without any doubt I can say that the metalic bonds </span><span> allows metals to conduct electricity. I'm pretty sure it will help you! Regards.</span>
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What two things effect the amount of kinetic energy an object has?
marta [7]

Answer:

mass and speed

Explanation:

the motion of an object depends on how fast it's travelling and also how much mass it has

3 0
3 years ago
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
What is the percent composition of hydrogen if a sample is found to contain 7 grams of hydrogen, 32 grams of nitrogen, and 7 gra
nata0808 [166]

<u>Answer:</u> The percent composition of hydrogen in the sample is 15.22 %

<u>Explanation:</u>

We are given:

Mass of hydrogen = 7 grams

Mass of nitrogen = 32 grams

Mass of carbon = 7 grams

Total mass of the sample = 7 + 32 + 7 = 46 grams

To calculate the percentage composition of hydrogen in sample, we use the equation:

\%\text{ composition of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of sample}}\times 100

Mass of sample = 46 g

Mass of hydrogen = 7 g

Putting values in above equation, we get:

\%\text{ composition of hydrogen}=\frac{7g}{46g}\times 100=15.22\%

Hence, the percent composition of hydrogen in the sample is 15.22 %

6 0
3 years ago
The first task you are assigned in your summer laboratory job is to prepare a concentrated naoh stock solution. the molecular we
tensa zangetsu [6.8K]

240 g NaOH

<em>Step 1</em>. Calculate the moles of NaOH

Moles of NaOH = 0.750 L solution × (8 mol NaOH/1 L solution) = 6 mol NaOH

Step 2. Calculate the mass of NaOH

Mass of NaOH = 6 mol NaOH × (40 g NaOH/1 mol NaOH) = 240 g NaOH

7 0
3 years ago
HELP ME OUT PLEASE!!!!!!!!
34kurt

Answer:

49.979

Explanation:

4 0
2 years ago
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