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Fantom [35]
3 years ago
12

For the reaction 2HNO3 + Mg(OH)2 → Mg(NO3) 2 + 2H2O, how many grams of magnesium nitrate are produced from 4 grams of nitric aci

d, HNO3 ?
Round your answer to the nearest tenth. Do not round your answers until the last step of the problem. If you do, the computer might mark your answer incorrect and you will be asked to complete your second trial for this assessment.

Use the following molar masses:

Element Molar Mass
Hydrogen 1
Magnesium 24
Nitrogen 14
Oxygen 16
Chemistry
1 answer:
kotegsom [21]3 years ago
8 0

Answer:

First check to make sure you have a balanced equation

Second use mole ratios to set up what you need to know

Finally convert the moles to grams using the correct molecular weight

I will get you started

According to the reaction above, one mole of Magnesium Nitrate is produced from 2 moles of Nitric Acid

1 mole of Mg(NO3)2/2 moles of HNO3

5 grams of Nitric acid contains

5 g * 1 mole/63 g of Nitric Acid = 0.079365079 moles

Explanation:

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The changing of a substance from a liquid into a vapor or gas is called
sergejj [24]

Answer: Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.

Explanation: Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.

6 0
1 year ago
In the presence of a base, blue litmus paper will .<br>O turn purple<br>stay blue<br>to turn red​
Inessa05 [86]

Answer:

In the presence of a base, blue litmus paper will turn red........

7 0
3 years ago
Read 2 more answers
What can you do differently for
drek231 [11]

Answer:

I would say, what helps me is really paying attention in class and asking questions, also making sure you study for upcoming test's and quizzes and completely assingments on time

Explanation:

3 0
3 years ago
How much water would need to be added to 1092 mL of a 54.7 M NaCl solution to make a 0.25 M solution?
babunello [35]

Answer:

237.8L of water would need to be added.

Explanation:

The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).

Then figure out what information you have and what is being found. In this case:

M1 = 54.7 M

V1 = 1092 mL = 1.092 L

M2 = 0.25 M

V2 = unknown

Then solve the equation for whatever you are trying to find.

M1V1=M2V2

V2=M1V1/M2

Now you need to plug everything in.

V2=(54.7M*1.091L)/0.25M

V2=238.93L

That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.

238.93L - 1.091L = 237.8L

Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.

I hope this helps.  Let me know if anything is unclear.

8 0
3 years ago
On the basis of the information above, a buffer with a pH = 9 can best be made by using
telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for \mathbf{H_3PO_4 , H_2PO^{-}_4 ,  HPO_4^{2-}} are \mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \  5\times 10^{-13}} respectively.

\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}

\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}

\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}

The reason while option D is the best answer is that, the value of pKa for both

\mathbf{H_2PO^{-}_4 ,\  \& \  HPO_4^{2-}} lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}

3 0
3 years ago
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