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Evgen [1.6K]
3 years ago
8

How much energy is transferred from a plant to an herbivore that eats it? A. 5% B. 10% C. 20% D. 90%

Chemistry
1 answer:
Tema [17]3 years ago
6 0

Answer:

B. 10%

Explanation:

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What is the molarity of a solution in which 25.3 grams of potassium bromide is dissolved in 150. mL of solution?
irga5000 [103]

Answer:

~1.417M

Explanation:

Molarity=(number of moles of solute)/(litres of solution)

In this case, we need to find moles of potassium bromide.

Mass=25.3g

Molar mass= 119g/mol

moles=(mass/molar mass)

        =(25.3)/(119)

        =0.2126moles of potassium bromide

Molarity=(0.2126)/(150/1000)

       ~1.417M

Hope this helps:)

6 0
3 years ago
Read 2 more answers
A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar. part a part complete how many milligrams of sugar does the m
kolezko [41]
Answer:
A 12.630 bar of chocolate contains 8315 milligrams of sugar

Explanation:
From the basics of conversion, we can find that:
1 gram is equal to 1000 mg
Using cross multiplication, we can find how many milligrams are present in 8.315 grams as follows:
1 gram .............> 1000 mg
8.315 grams ....> ?? mg
amount in milligrams = (8.315*1000) / 1 = 8315 milligrams

Hope this helps :)
5 0
3 years ago
As the population in a small town grows, a new water treatment plant is built. The plant is built on a local river where it can
xenn [34]

Answer:

A. Treated water from the plant would affect communities downriver.

Explanation:

7 0
4 years ago
Please help me, Thank you!
babymother [125]

Answer:

amount of charge

Explanation:

Oxygen and sulfur are both in Group 16, which means they have a -2 charge. They have two more electrons than protons, making the charge of the ion negative.

Hope that helps.

4 0
3 years ago
Read 2 more answers
A sample of an ionic compound NaA, where A- is the anion of a
vitfil [10]

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

6 0
3 years ago
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