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jonny [76]
2 years ago
5

If 200mL of 0.60 M MgCl2 (aq) is added to 400mL of distilled water, what is the concentration of Mg2+(aq) in the resulting solut

ion with the total volume of 600mL? A.0.20M. B.0.30M. C.0.60M D.1.2M
Chemistry
1 answer:
lara [203]2 years ago
3 0

Answer:

A.0.20M

Explanation:

c 1 V 1 = c 2 V 2

Initial Volume, V1 = 200 mL

Final Volume, V2 = 200 + 400 = 600 mL

Initial Concentration, c1 = 0.60 M

Final Concentration, c2= ?

Solving for c2;

c2 = c1v1 / v2

c2 = 0.60 * 200 / 600

c2 = 0.20M

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butalik [34]

Water boils at 100 Degrees Celsius or 212 degrees Fahrenheit

8 0
3 years ago
4. A 0.51 kg solution contains 87 mg of potassium iodide. Calculate the W/W concentration
anzhelika [568]

Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.

<h3>Definition of percent composition </h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.

<h3>Percentage Composition in this case</h3>

In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.

Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

percentage composition of potassium iodide=\frac{87 mg}{510000 mg} x100

<u><em>percentage composition of potassium iodide= 0.017%</em></u>

Finally, the percent composition of potassium iodide in this sample is 0.017%.

Learn more about percent composition:

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3 0
2 years ago
An insect population increases and then decreases as the food supply changes.
Arada [10]

Answer:

self regulation

Explanation:

4 0
3 years ago
Read 2 more answers
Which elements form are ionic compounds check all the boxes that apply
morpeh [17]

Answer/Explanation:

Aluminum and oxygen

Fluorine and oxygen

Ionic compounds are formed when any type of metal is combined with a non-metal such as carbon, nitrogen, oxygen, sulfur, phosphorus, and selenium.

(any)metal + nonmetal = ionic compound

7 0
3 years ago
How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?
BartSMP [9]

Answer:

m_{AgNO_3}=577.6mg

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:

n_{LiOH}=81.5mg*\frac{1g}{1000mg}*\frac{1mol}{23.95g}  =0.0034molLiOH

Thus, since there is a 1:1 mole ratio between lithium hydroxide and silver nitrate (169.87 g/mol) the resulting milligrams turn out to be:

m_{ AgNO_3}=0.0034molLiOH*\frac{1molAgNO_3}{1molLiOH} *\frac{169.87gAgNO_3}{1molAgNO_3} *\frac{1000gAgNO_3}{1gAgNO_3} \\\\m_{AgNO_3}=577.6mg

Best regards!

7 0
2 years ago
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