The steps leading to an algal-bloom fish kill

Chemistry

1 answer:

adoni [48]3 years ago

3 0

Answer:

the bacteria increase in number and use up the dissolved oxygen in the water. When the dissolved oxygen content decreases, many fish and aquatic insects cannot survive. This results in a dead area. Algal blooms may also be of concern as some species of algae produce neurotoxins.Explanation:

You might be interested in

Help pls :) I am stuck on this chemistry question about percentage yields!

Charra [1.4K]

You just switch them around

4 0

3 years ago

What is Industrial chemistry?

mel-nik [20]

Answer:

study of chemical used in industries

6 0

3 years ago

If you start with 6 moles of N2 and 6 moles of H2 (meaning you won't have enough of 1 of the ingredients), how many moles of NH3

Ivahew [28]

Answer:

$4molNH_3$

Explanation:

Hello there!

In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:

$N_2+3H_3\rightarrow 2NH_3$

We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:

$n_{NH_3}=6molN_2*\frac{2molNH_3}{1molN_2}=12molNH_3\\\\ n_{NH_3}=6molH_2*\frac{2molNH_3}{3molH_2}=4molNH_3\\$

Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.

Regards!

8 0

3 years ago

For the balanced equation shown below, if the reaction of 0.112 grams of

Alekssandra [29.7K]

Answer:

The answer is 74.5%.

Explanation:

As we know that % yield= $\frac{actual yield}{theoretical yield}$ x 100%.

Therefore,

Step 1 Calculate Theoretical yield:

0.112 $H_{2}$ x $\frac{1 mol H_{2} }{2.016 g of H{2} }$ x $\frac{4 mol H_{2}O }{4 mol H_{2} }$ x $\frac{18.02 g H_{2}O }{1 mol H_{2}O}$ = 1.001 g $H_{2}O$