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forsale [732]
2 years ago
5

Given the following thermochemical equation 2N2O(g) → 2N2(g) + O2(g) ∆Hº = –166.7 kJ/mol find the amount of heat that will be pr

oduced when a 2.25 g sample of N2O(g) decomposes into N2 and O2.A. 8.51 kJB. 4.25 kJC. 74.1 kJD. 2.13 kJE. 37.1 kJ
Chemistry
1 answer:
soldi70 [24.7K]2 years ago
5 0

Answer:

Explanation:

2N₂O(g) → 2N₂(g) + O₂(g)

molecular weight of N₂O = 44

∆Hº = –166.7 kJ/mol

44 g of N₂O decomposes to give 166.7 kJ of heat

2.25 g of N₂O decomposes to give 166.7 x 2.25 / 44 kJ of heat

= 8.51 kJ of heat .

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Carbon-14 has a half-life of 5,730 years. In how many years will 120 grams of Carbon-14 decay to 15 grams? 11,460 years 17,190 y
erastova [34]
120 grams of Carbon-14 decays to 15 grams in 17,190 years.
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3 years ago
Arrange these elements based on their atomic radii. ga, f, s, as
mamaluj [8]

please have look at Periodic table , you will solve it yourself !

5 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
What allows nutrients to enter Plants and animals are able to be transported
elena55 [62]
I think it’s the cell membrane if you’re talking about animal cells and plant cells.
3 0
3 years ago
2.
katovenus [111]

Answer:

A. 6atm

Explanation:

Using pressure law equation:

P1/T1 = P2/T2

Where;

P1 = initial pressure (atm)

T1 = initial temperature (K)

P2 = final pressure (atm)

T2 = final temperature (K)

According to this question;

P1 = 3 atm

P2 = ?

T1 = 120K

T2 = 240K

Using P1/T1 = P2/T2

3/120 = P2/240

Cross multiply

240 × 3 = P2 × 120

720 = 120P2

P2 = 720/120

P2 = 6atm

4 0
2 years ago
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