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3 years ago

Magnesium hydroxide (Mg(OH)2): g/mol

2 answers:

8 0

To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.

Mg = 24.305 x 1 = 24.305 amu

O = 15.9994 x 2 =31.9988 amu

H = 1.0079 x 2 = 2.0158 amu

Then, add all the components to get the atomic mass of the molecule.

24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu

The atomic mass is just equivalent to its molar mass.

So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.

Bumek [7]3 years ago

6 0

Answer:
molar mass of Mg(OH)2 = 58.305 grams/mole

Explanation:
From the periodic table:
atomic mass of Mg = 24.305 grams
atomic mass of oxygen = 16 grams
atomic mass of hydrogen = 1 gram

Now the given compound has:
1 mole of Mg, 2 moles of oxygen and 2 moles of hydrogen
Therefore:
molar mass of Mg(OH)2 = 24.305 + 2(16) + 2(1) = 58.305 grams/mole

You might be interested in

A water molecule is composed of a large oxygen atom bonded to two small hydrogen atoms. Which of the following would make the be

sleet_krkn [62]

The best and most correct answer among the choices provided by your question is the third choice or letter C.

The best model of a water molecule would be: Two small, plastic balls attached to a larger plastic ball by toothpicks

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

3 years ago

Read 2 more answers

Can someone please help me with my science research I kind of need help in a few things that I listed, I’m currently working on

dalvyx [7]

Answer:

For your first question, Curium does not occur naturally on Earth, meaning that it is not produced naturally on Earth. However, it can be formed in nuclear reactors.

For your second question, Curium has been used to provide power to electrical equipment used on space missions, but doesn't seem to be that important overall.

Explanation:

Hope this helped!

7 0

2 years ago

A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

(B) is correct.

4 0

3 years ago

What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water

omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

, in calcium chloride, This will help you calculate the mass of chloride anions,

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

mole of calcium chloride contains

moles of chlorine atoms.

35.453

g mol

−

110.98

g mol

−

⋅

100

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

−

This implies that your sample contains

0.543

g CaCl

⋅

63.89 g Cl

−

100

g CaCl

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

⋅

1 mole CaCl

110.98

0.004893 moles CaCl

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

mL solution

⋅

0.004893 moles CaCl

500

mL solution

0.009786 moles CaCl

You can thus say your solution has

[

CaCl

]

0.009786 mol L

−

Since every mole of calcium chloride delivers

moles of chloride anions to the solution, you can say that you have

[

−

]

⋅

0.009786 mol L

−

[

−

]

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

mL solution

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

mole Cl

−

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

0.0341 g/100 mL

0.0341% m/v

because you have

1 L

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

−

]

0.01957 mol L

−

you have

⋅

0.01957 mol

⋅

−

⋅

500

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

Explanation:

i think this is it

8 0

2 years ago

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jasenka [17]

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8 0

3 years ago