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valentinak56 [21]
3 years ago
6

Which of the following are some examples of rapid environmental changes?

Chemistry
2 answers:
Zina [86]3 years ago
6 0
I think it’s weather
Marysya12 [62]3 years ago
6 0

Answer: Fires, Weather and Landslides are examples of rapid environmental changes

Explanation:

Environmental changes are currently one of the world’s major issues. Rapid changes in the environment is mostly attributed to anthropogenic factors in the environment such as CO2 emission, gas combustion in industries soil erosion from man-made inputs leading to landslides from road networks, hence the changes in the weather condition as well as fire, and landslides. As at 40 – 60 million years ago during the cretacious period, the earth experienced warmer conditions conducive enough for habitation, whereas existing conditions now in the 21st century are colder, producing harsh conditions that are sometimes considered to be bad to breath in. Hence the reason why Fires arising from deforestation, landslides from bad man-made planning in constructional projects and Climate change causing a depletion in the Ozone layer are considered to be an example of rapid environmental changes.

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4+4=?
lesya692 [45]

Answer:

8

Explanation:

3 0
3 years ago
Read 2 more answers
An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average
Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = \frac{Displacement }{Change in time} = Δx / Δt = \frac{x2 - x1}{t2 - t1}

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{2100km }{1000km/h}

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{1300km }{800km/h}

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = \frac{x2 - x1}{t2 - t1}

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = \frac{1300 - 2100}{1.625 - 2.1}

Average velocity = 1684.21 km/h

7 0
3 years ago
The ka of hypochlorous acid (hclo) is 3.0 ⋅ 10−8 at 25.0 °c. calculate the ph of a 0.0375m hypochlorous acid solution.
Scrat [10]
We can set up an ICE table for the reaction:                      
                      HClO          H+     ClO-
Initial              0.0375       0        0
Change         -x               +x      +x
Equilibrium    0.0375-x     x        x

We calculate [H+] from Ka:     
     Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)

Approximating that x is negligible compared to 0.0375 simplifies the equation to         
     3.0x10^-8 = (x)(x)/0.0375     
     3.0x10^-8 = x2/0.0375     
     x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9     
     x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.

We can now calculate pH:     
     pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
6 0
3 years ago
I need all solutions
marshall27 [118]

Answer:

Answer is in the attachment

6 0
2 years ago
4. a compound called pyrene has the empirical formula c8h5. when 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boilin
e-lub [12.9K]

The molecular mass of pyrene is 204.4 g/mol.

From;

ΔT = Kb m i

Where;

  • ΔT = boiling point elevation
  • Kb = boiling point constant
  • m = molality
  • i = Van't Hoff factor

Since the compound is molecular; i = 1

The number of moles of pyrene = 4.04 g/MM

Where; MM = molar mass of pyrene

molality = number of moles of pyrene/mass of solvent in Kg

The mass of solvent = 10 g or 0.01 Kg

molality =  4.04 g/MM/0.01

ΔT = Boiling point of solution - Boiling point of pure solvent

ΔT = 85.1°C - 80.1°C

ΔT = 5°C

5 = 2.53 × 4.04 g/MM/0.01 × 1

5 = 10.22 × 1/0.01 MM

0.05MM = 10.22

MM= 10.22/0.05

MM= 204.4 g/mol

Learn more: brainly.com/question/2292439

3 0
2 years ago
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