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photoshop1234 [79]
2 years ago
9

Xà phòng hoá hoàn toàn 22,2g hỗn hợp gồm 2 este HCOOC2h5 và ch3cooch3 bằng dung dịch naoh 1m ( đun nóng) thể tích dung dịch naoh

tối thiểu cần dùng là
Chemistry
1 answer:
lisabon 2012 [21]2 years ago
4 0

Answer:

I can't understand your language? can you write in English

I'm sorry

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A solution is prepared by dissolving 8.75 grams of sodium chloride in enough volume of water to produce 112.0 grams of the NaCl
ICE Princess25 [194]

Answer : The percent by mass of the solution is 7.81 %

Explanation : Given,

Mass of NaCl = 8.75 g

Mass of solution = 112.0 g

Now we have to determine the percent by mass of the solution.

\text{Mass percent}=\frac{\text{Mass of NaCl}}{\text{Mass of solution}}\times 100

Putting values in above equation, we get:

\text{Mass percent}=\frac{8.75g}{112.0g}\times 100=7.81\%

Therefore, the percent by mass of the solution is 7.81 %

6 0
3 years ago
Chemical formula for Potassium Hydrogen Nitrate
My name is Ann [436]

Answer:

KNO3

Explanation:

8 0
3 years ago
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
A sample originally contains 24 grams of a radioactive isotope. After 18 days, 18.8 grams of the isotope remain in the sample. W
LenKa [72]
Around 36 or 37 days, because of the decimal
8 0
2 years ago
B. 10% is the correct answer
GarryVolchara [31]
No the right answer is c
5 0
2 years ago
Read 2 more answers
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