analyzing data
proportional relationships in chemical reactions
a chemist runs three trials of an experiment to produce ammonia (nh₃) from nitrogen (n₂) and hydrogen (h₂). the unbalanced equation is
n₂ + h₂ → nh₃
the chemist places different amounts of the reactants in a reaction vessel, adjusts some of the conditions such as pressure and temperature, and then records the amounts of the reactants and the products in the table.
amount of products and reactants for n₂ + h₂ → nh₃
| trial | original amount (n₂ mol) | original amount (h₂ mol) | original amount (nh₃ mol) | final amount (n₂ mol) | final amount (h₂ mol) | final amount (nh₃ mol) | change in amount (n₂ mol) | change in amount (h₂ mol) | change in amount (nh₃ mol) |
|-------|--------------------------|--------------------------|---------------------------|-----------------------|-----------------------|---------------------------|---------------------------|---------------------------|----------------------------|
| 1 | 0.345 | 0.345 | 0.000 | 0.335 | 0.315 | 0.020 | | | |
| 2 | 0.102 | 0.345 | 0.000 | 0.032 | 0.135 | 0.140 | | | |
| 3 | 0.345 | 0.298 | 0.000 | 0.305 | 0.178 | 0.080 | | | |
1. sep use mathematics calculate (in moles) the amount of each reactant that was used up in the reaction and the amount of product that was produced. record your answers in the table.
enter your answer here.

Question

analyzing data
proportional relationships in chemical reactions
a chemist runs three trials of an experiment to produce ammonia (nh₃) from nitrogen (n₂) and hydrogen (h₂). the unbalanced equation is
n₂ + h₂ → nh₃
the chemist places different amounts of the reactants in a reaction vessel, adjusts some of the conditions such as pressure and temperature, and then records the amounts of the reactants and the products in the table.
amount of products and reactants for n₂ + h₂ → nh₃
| trial | original amount (n₂ mol) | original amount (h₂ mol) | original amount (nh₃ mol) | final amount (n₂ mol) | final amount (h₂ mol) | final amount (nh₃ mol) | change in amount (n₂ mol) | change in amount (h₂ mol) | change in amount (nh₃ mol) |
|-------|--------------------------|--------------------------|---------------------------|-----------------------|-----------------------|---------------------------|---------------------------|---------------------------|----------------------------|
| 1     | 0.345                    | 0.345                    | 0.000                     | 0.335                 | 0.315                 | 0.020                     |                           |                           |                            |
| 2     | 0.102                    | 0.345                    | 0.000                     | 0.032                 | 0.135                 | 0.140                     |                           |                           |                            |
| 3     | 0.345                    | 0.298                    | 0.000                     | 0.305                 | 0.178                 | 0.080                     |                           |                           |                            |
1. sep use mathematics calculate (in moles) the amount of each reactant that was used up in the reaction and the amount of product that was produced. record your answers in the table.
enter your answer here.

Accepted Answer

# Explanation:
## Trial 1: Calculate change in $ N_2 $
The change in amount of $ N_2 $ is original amount minus final amount.
$ 0.345 - 0.335 = 0.010 $ mol

## Trial 1: Calculate change in $ H_2 $
Change in $ H_2 $ is original minus final.
$ 0.345 - 0.315 = 0.030 $ mol

## Trial 1: Calculate change in $ NH_3 $
Change in $ NH_3 $ is final minus original (original is 0).
$ 0.020 - 0.000 = 0.020 $ mol

## Trial 2: Calculate change in $ N_2 $
$ 0.102 - 0.032 = 0.070 $ mol

## Trial 2: Calculate change in $ H_2 $
$ 0.345 - 0.135 = 0.210 $ mol

## Trial 2: Calculate change in $ NH_3 $
$ 0.140 - 0.000 = 0.140 $ mol

## Trial 3: Calculate change in $ N_2 $
$ 0.345 - 0.305 = 0.040 $ mol

## Trial 3: Calculate change in $ H_2 $
$ 0.298 - 0.178 = 0.120 $ mol

## Trial 3: Calculate change in $ NH_3 $
$ 0.080 - 0.000 = 0.080 $ mol

# Filling the table (rows as trials, columns as changes):
| Trial | Change in $ N_2 $ (mol) | Change in $ H_2 $ (mol) | Change in $ NH_3 $ (mol) |
|-------|---------------------------|---------------------------|----------------------------|
| 1     | 0.010                     | 0.030                     | 0.020                      |
| 2     | 0.070                     | 0.210                     | 0.140                      |
| 3     | 0.040                     | 0.120                     | 0.080                      |

# Answer:
The table with changes is as above. For each trial:
- Trial 1: $ N_2 $: 0.010, $ H_2 $: 0.030, $ NH_3 $: 0.020
- Trial 2: $ N_2 $: 0.070, $ H_2 $: 0.210, $ NH_3 $: 0.140
- Trial 3: $ N_2 $: 0.040, $ H_2 $: 0.120, $ NH_3 $: 0.080

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QUESTION IMAGE

Question

Explanation:

Trial 1: Calculate change in \( N_2 \)

Trial 1: Calculate change in \( H_2 \)

Trial 1: Calculate change in \( NH_3 \)

Trial 2: Calculate change in \( N_2 \)

Trial 2: Calculate change in \( H_2 \)

Trial 2: Calculate change in \( NH_3 \)

Trial 3: Calculate change in \( N_2 \)

Trial 3: Calculate change in \( H_2 \)

Trial 3: Calculate change in \( NH_3 \)

Filling the table (rows as trials, columns as changes):

Answer:

trial	original amount (n₂ mol)	original amount (h₂ mol)	original amount (nh₃ mol)	final amount (n₂ mol)	final amount (h₂ mol)	final amount (nh₃ mol)	change in amount (n₂ mol)	change in amount (h₂ mol)	change in amount (nh₃ mol)
2	0.102	0.345	0.000	0.032	0.135	0.140
3	0.345	0.298	0.000	0.305	0.178	0.080