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Galina-37 [17]
3 years ago
15

If an area bans pesticides and fertilizers to clean up the water supply, how might this affect farmers or food resources in that

area?
Chemistry
1 answer:
Lelechka [254]3 years ago
7 0

Answer:

Farmers will be negatively affected, there will not be as many food resources.

Explanation:

If an area bans pesticides and fertilizers, farmers will have a difficult time growing crops. If farmers can't grow crops, the amount of food resources will decrease.

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From the balanced reaction:
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Answer:

The mass of O₂ that will be needed to burn 36.1 g B₂H₆ is 125.29 g.

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How many valence electrons does chlorine have?
Advocard [28]
Valence Electrons has \boxed{7} chlorine
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What are the products obtained in the electrolysis of molten nai?
yanalaym [24]
Answer is: sodium (Na) and iodine (I₂).

<span> First ionic bonds in this salt are separeted because of heat: 
</span>NaI(l) → Na⁺(l) + I⁻(l).

Reaction of reduction at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.

2Na⁺(l) + 2e⁻ → 2Na(l).

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The anode is positive and the cathode is negative.


4 0
3 years ago
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
LiRa [457]

Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
3 years ago
How many significant figures in 8400
IrinaK [193]

Answer:

there are two significant figures is the number 8400

Explanation:

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