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goldenfox [79]
3 years ago
15

What happens if I overshoot the endpoint in titration?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
5 0
Your solution in the conical flask will be acidic since you will have gone beyond the amount of acid that should completely neutralise the base completely. thax
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How many grams of h3po4 are in 255 ml of a 4.50 m solution of h3po4?
murzikaleks [220]

H3PO4 has molecular weight of approximately 98 grams per mole. 4.50 M is equal to 4.50 mole per 1000 mL solution of H3PO4. 255 mL times 4.50 mol /1000 mL times 98 g/mol is equal to 112.455 grams. Note that I automatically equate 1 Liter to 1000 mL since the given volume is in mL for easier computation.

7 0
2 years ago
How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2
Nuetrik [128]

Answer:

Moles of ScCl_3 = 6 moles

Explanation:

The reaction of Sc and Cl_2 to make ScCl_3 is:

2Sc+3Cl_2⇒2ScCl_3

The above reaction shows that 2 moles of Sc  can react with 3 moles of Cl_2 to form ScCl_3.

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of Cl_2 = Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}

Moles of Cl_2 = 10 *\frac{3 moles of Cl_2}{2 Moles of Sc}

Moles of Cl_2 =15 moles

So 15 moles of Cl_2 are required to react with 10 moles of Sc but we have 9 moles of Cl_2 , it means Cl_2 is limiting reactant.

Moles of ScCl_3=Given\  Moles\  of\ Cl_2 *\frac{2\  Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}

Moles\ of\  ScCl_3=9 *\frac{2\  Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}

Moles of ScCl_3= 6 moles

4 0
2 years ago
Read 2 more answers
What is the electronic structure of carbon
horsena [70]

Answer:

[He] 2s2 2p2

Explanation:

5 0
3 years ago
Information-
tekilochka [14]

Answer:

So first thing to do in these types of problems is write out your chemical reaction and balance it:

Mg + O2 --> MgO

Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.

To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.

The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.

The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.

Percent yield is acutal/theoretical, .66/.693, or 95.24%.

I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.

Hope this helps.

6 0
2 years ago
How is stoicheometry used to <br>keep camels alive​
Novay_Z [31]

Answer:

by heart beat camel lives

8 0
2 years ago
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