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lara31 [8.8K]
2 years ago
15

Which of the following represents an endothermic reaction?

Chemistry
1 answer:
Softa [21]2 years ago
8 0

Answer:

<h2><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em></h2>

The second one would be one of the application of an endothermic reaction as because endothermic reaction always have the heat in the process to mix the reactants and form the products as here in a burning wood when the wood is supplied with heat then it change into products as ashes and smoke fumes So here due to the heat is involved to form the products thats the reaction is endothermic reaction.

Hope it helps

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Liula [17]

I believe that #1 is the lie, but I'm not great at this subject.

4 0
3 years ago
Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
nignag [31]

Answer:  1) Maximum mass of ammonia  198.57g  

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of  the excess Reactant, that means the H2 with 3,44g

Explanation:

  • In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

  • Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the  molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶  38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

  • As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element  that would be completey consumed, and the maximum mass of ammonia will be produced from it.
  • We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66  mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

  • In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of  H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

3 0
3 years ago
Which agent of erosion can create a limestone cave?
Ivanshal [37]
Water containing carbonic acid and calcium
5 0
3 years ago
If the volume of the original sample in Part A ( P1 = 242 torr , V1 = 27.0 L ) changes to 80.0 L , without a change in the tempe
Brrunno [24]

The new pressure is 81.675 torr

Since temperature and moles are held constant, we use Boyle's Law:

A gas law known as Boyle's law asserts that a gas's pressure is inversely proportional to its volume when it is held at a fixed temperature and of a given mass.

To put it another way, as long as the temperature and volume of the gas remain constant, the pressure and volume of the gas are inversely proportional to one another.

The Anglo-Irish chemist Robert Boyle proposed Boyle's law in the year 1662.

P1V1=P2V2. Simply plug in your values. The units can remain in torr. Converting to atmospheres is not needed.

(242 torr)(27.0 L)=P2(80.0 L)

P2=[(242)(27)]/80 = 81.675 torr

Hence The new pressure is 81.675 torr

Learn more about Boyle's Law here

brainly.com/question/26040104

#SPJ4

3 0
1 year ago
A 7.00-mL portion of 8.00 M stock solution is to be diluted to 0.800 M. What will be the final volume after dilution? Enter your
amm1812

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically,         Molarity = \frac{\text{no. of moles}}{\text{Volume in liter}}

Also, when number of moles are equal in a solution then the formula will be as follows.

                     M_{1} \times V_{1} = M_{2} \times V_{2}

It is given that M_{1} is 8.00 M, V_{1} is 7.00 mL, and M_{2} is 0.80 M.

Hence, calculate the value of V_{2} using above formula as follows.

                    M_{1} \times V_{1} = M_{2} \times V_{2}

                 8.00 M \times 7.00 mL = 0.80 M \times V_{2}

                      V_{2} = \frac{56 M. mL}{0.80 M}

                                  = 70 ml

Thus, we can conclude that the volume after dilution is 70 ml.

7 0
3 years ago
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