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Masteriza [31]
3 years ago
9

in times of hot dry weather some cities ban the use of garden sprinklers why do you think there is such a rule

Chemistry
2 answers:
Over [174]3 years ago
8 0

Answer:

To restrict the amount of water being utilized from a limited supply.

Explanation:

A sprinkler or a hosepipe ban gets imposed in certain cities during hot weather conditions in order to conserve the declining levels of drinking water, that is, stored in the reservoirs at the time of drought. Once the rainy season comes back the ban gets lifted as the reservoirs get filled once again, and it is more essential to provide water to the public, as the grass can always grow again.

Solnce55 [7]3 years ago
7 0

ban the use of garden sprinklers during hot dry weather because If water use doesn't decrease soon, pressure could start dropping and supplies could cut out without warning. Weeks of dry weather, high temperatures and warm winds have pushed water use almost to the limit.


 

                                                                                                                                      

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Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

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Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

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Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

5 0
4 years ago
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g Suppo
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Answer:

[SO_3]=0.25M

Explanation:

Hello there!

In this case, since the integrated rate law for a second-order reaction is:

[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}

Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:

[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\

[SO_3]=0.25M

Best regards!

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8 0
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