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Snowcat [4.5K]
3 years ago
7

Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of Cl2 loss is 4.44 × 1

0-2 M/s, what is the rate of formation of NOCl?2 NO(g) + Cl2(g) → 2 NOCl(g)1.11 × 10-1 M/s2.22 × 10-2 M/s8.88 × 10-2 M/s1.61 × 10-2 M/s4.44 × 10-2 M/s
Chemistry
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

8.88 x 10⁻² M/s

Explanation:

The rate  of reaction for:

                    NO(g) + Cl₂ (g) ⇒  2NOCl(g)

is rate = -ΔNO/Δt = -ΔCl2/Δt = 1/2 ΔNOCl/Δt

so  ΔNOCl/Δt = 2 ΔCl2/Δt  = 2  x 4.44 × 10⁻²  M/s = 8.88 x 10⁻² M/s

In general given a reaction

                            aA + bB ⇒ cC + dD

rate = -1/a ΔA/Δt = -1/b ΔB/Δt = 1/c ΔC/Δt = 1/d ΔD/Δt

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What volume does 2.25g of nitrogen gas, N2, occupy at 273 Celsius and 1.02 atm​
kotykmax [81]
<h2><u>Answer:</u></h2>

0.126 Liters

<h2><u>Explanation:</u></h2>

V = mRT / mmP

First, convert the 2.25g of Nitrogen gas into moles. (m in the equation above)

2.25g x 1 mole / 28.0g = 0.08036 moles = m

28.0g = mm

Next, convert the 273 Celsius into Kelvin. (T in the equation above)

273 Celsius + 273.15 = 546.15K = T

R = 0.08206L*atm/mol*K

(Quick Note: The R changes depending on the Pressure Unit so do not use this number every time.)

Now, plug everything into the equation.

V = (0.08036)(0.08206)(546.15)/(28.0)(1.02)

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5 0
3 years ago
HELP PLEASE THE OTHER 'ANSWER' ISNT EVEN AN ANSWER!
hodyreva [135]

Answer:

most likely that (2) the replicated experiment was performed incorrectly.

Why, u ask? u dare question me:

1- The initial experiment invalidness cannot be proven.

2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><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><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>

3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.

4- Already knowing the data and errors would increase the precision of the replicated experiment.

5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.

happy learning!

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