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mylen [45]
3 years ago
12

Increased irrigation demands: decrease the rate of transpiration increase the rate of surface water evaporation change the amoun

t of precipitation in an area do not affect the water cycle
Chemistry
1 answer:
drek231 [11]3 years ago
7 0

Answer: Increase the rate of surface water evaporation

Irrigation is a process, in which the crops are supplied with water, to ensure their proper growth. Increased in the irrigation supply can be because of the environmental conditions like hot summer season or may be because of the type of soil is semi arid or arid means the soil does not retains moisture efficiently. This will result in increase in the rate of evaporation of the surface water.

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HELP ME ON NUMBER 8 ASAP
Varvara68 [4.7K]

Answer:

2. hope it helps. please follow me I'm new here

3 0
3 years ago
You calculate that your semester average in history is 97.5. When you get your report card, your average is 96. What was the per
andrew11 [14]
Ok percent error is abs(calculated-actual)/actual(100%)

So 1.5/96 *100%
5 0
3 years ago
Read 2 more answers
Looking at the following equation, what coefficient should be placed in front of the H2? Fe + H2SO4 --> Fe2(SO4)3 + H2
Hatshy [7]

Answer:

3 (three)

Explanation:

2 Fe + 3H2SO4 = Fe2(SO4)3 + 3 H2 (basically just balance both sides)

6 0
3 years ago
A sample of monochloro pentaborane (9), b5h8cl, has a mass of 20.1 g. what amount, in moles, does this mass represent?
dolphi86 [110]

The number of moles in monochloro pentaborane (9) B_5H_8Cl is 0.37056.

We know that ,

Number of moles =   Given mass/ Molar mass

Given mass of monochloro pentaborane (9) is 20.1 g

The molar mass of monochloro pentaborane (9) is

                                        =  5*B +  8*H + Cl

                                        =  5* 10.811 + 8 * 1 + 35.453

                                        = 54.241

Thus number of moles =  20.1/54.241

                                      = 0.37056

To know more about mole concept

brainly.com/question/20483253

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5 0
2 years ago
The buffer solution is used to control the pH to insure that it does not become too high because excessively basic solutions cou
oksian1 [2.3K]

Answer:

pH = 12.7

Explanation:

First, we have to calculate the [Ca²⁺] in a solution of about 250 ppm CaCO₃.

\frac{250mgCaCO_{3}}{L} .\frac{1gCaCO_{3}}{1000mgCaCO_{3}} .\frac{1molCa^{2+} }{100gCaCO_{3}} =2.5 \times 10^{-3} M

Now, let's consider the dissolution of Ca(OH)₂ in water.

Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)

The solubility product Ksp is:

Ksp = [Ca²⁺] × [OH⁻]²

[OH⁻] = √(Ksp/[Ca²⁺]) = √(6.5 × 10⁻⁶/2.5 × 10⁻³) = 5.1 × 10⁻² M

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (5.1 × 10⁻²) = 1.3

pH + pOH = 14 ⇒ pH = 14 - pOH = 14 - 1.3 = 12.7

5 0
3 years ago
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