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Leto [7]
3 years ago
13

Which of the following developments would most likely occur if average summer air temperatures in the Arctic increased by 4°C?

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

A. Increase in permafrost thickness

Explanation:

<u>Climate warming (like of 4°C) would increase the active layer thickness (ALT) of permafrost</u>, which can cause changes in the hydrological cycle, ecological processes, and carbon flux in cold regions.

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In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.
mihalych1998 [28]

In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol t_{1/2}.

(a) For first order reaction, rate constant and half life time are related to each other as follows:

k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}

Thus, rate constant of the reaction is 0.066 min^{-1}.

(b) Rate equation for first order reaction is as follows:

k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}

now, 75% of the compound is decomposed, if initial concentration [A_{0} ] is 100 then concentration at time t [A_{t} ] will be 100-75=25.

Putting the values,

0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)

On rearranging,

t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min

Thus, time required for 75% decomposition is 21 min.

8 0
4 years ago
A 35,000 Newton car runs up a hill that is 45.5 meters high in 1.6 seconds. what power did it exert?
Evgesh-ka [11]

Answer:

995.313KW

Explanation:

the explanation is in the picture

please like and Mark as brainliest

7 0
3 years ago
If you had a pot of boiling water that is loosing volume is that at equilibrium? What are some things you can do to a pot of boi
ki77a [65]

Answer:

are you rich in points brother ? or sister ?

8 0
3 years ago
The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval
worty [1.4K]
<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t  

The average rate of disappearance of ozone ... is found to  

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t  

rate = -(1/2)*(-7.25 × 10^–3 atm)

      = 3.625 × 10^–3 atm  

Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

              = 1.0875 x 10-2 atm over the same time interval

4 0
3 years ago
Please help, I really don’t understand this!!!
kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen  = 32 / 30

Mass of Oxygen = 1.06 grams

5 0
3 years ago
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