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Virty [35]
3 years ago
7

What is the concentration of magnesium bromide, in ppm, if 133.4 g MgBr2 dissolved in 1.84 L water. Then solve for the bromine c

oncentration in ppm
Chemistry
1 answer:
Serhud [2]3 years ago
7 0

Answer: 0.0725ppm

Explanation:

133.4g of MgBr2 dissolves in 1.84L of water.

Therefore Xg of MgBr2 will dissolve in 1L of water. i.e

Xg of MgBr2 = 133.4/1.84 = 72.5g

The concentration of MgBr2 is 72.5g/L = 0.0725mg/L

Recall,

1mg/L = 1ppm

Therefore, 0.0725mg/L = 0.0725ppm

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boyakko [2]

Answer:

SO SORRY IF THIS IS WRONG BUT I HOPE THIS HELPS

Explanation:

Primary fertilizers include substances derived from nitrogen, phosphorus, and potassium. Various raw materials are used to produce these compounds. ... The phosphorus component is made using sulfur, coal, and phosphate rock. The potassium source comes from potassium chloride, a primary component of potash.

3 0
3 years ago
Read 2 more answers
How can air pollution affect plants?
GarryVolchara [31]

Answer:

C. air pollution absorbs carbon dioxide

Explanation:

Carbon dioxide's role in the greenhouse effect is a major contributor to air pollution. Radiation and heat emanating from the earth's surface need to be released out into the atmosphere. But because carbon dioxide levels are so high, there is an ozone effect on the ground level.

8 0
2 years ago
scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14
Mars2501 [29]

 The much  of the sample that would remain  unchanged  after 140 seconds is 2.813 g

Explanation

Half life  is time taken for the quantity  to reduce  to half its original value.

if the half life  for Scandium  is 35 sec, then the number  of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

               22.5 g after second half life = 22.5 x 1/2 =11.25 g

            11.25 g after third half life = 11.25 x 1/2 = 5.625 g

             5.625 after  fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g  of Scandium 47 remains  unchanged.

4 0
2 years ago
Which of the following would require the largest volume of 0.100 M sodium hydroxide solution for neutralization?:
bazaltina [42]

Answer:10.0 mL of 0.00500 M phosphoric acid

Explanation:

If we look at the Ka values of the acids, we will realize that phosphoric acid has a Ka of 7.1 * 10-3. It is the only acid in the list having acid dissociation constant less than 1. This means that it does not ionize easily in solution and a very large volume of base must be added to ensure that it reacts completely. Acids with Ka >1 are generally regarded as strong acids. All the acids listed have Ka>1 except phosphoric acid.

8 0
3 years ago
How much heat energy is produced by 0.5 Wh of electrical energy
n200080 [17]
1 W is equivalent to 1 J/s

So,
0.5 Wh = 0.5 (J/s) (h)

Converting h to s,
0.5 (J/s) (h) (60min/1h) (60s/1min) = 1800J

Thefore,
0.5Wh is 1800J or 1.8kJ of heat energy

6 0
3 years ago
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