the answer is 0.000097 KM
Answer:
7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.
Explanation:
Let the volume of 10% acid solution used to make the mixture = x L
So, the volume of 30% acid solution used to make the mixture = y L
Total volume of the mixture = <u>x + y = 30 L .................. (1)
</u>
For 10% acid solution:
C₁ = 10% , V₁ = x L
For 30% acid solution :
C₂ = 30% , V₂ = y L
For the resultant solution of sulfuric acid:
C₃ = 25% , V₃ = 30 L
Using
C₁V₁ + C₂V₂ = C₃V₃
10×x + 30×y = 25×30
So,
<u>x + 3y = 75 .................. (2)
</u>
Solving 1 and 2 we get,
<u>x = 7.5 L
</u>
<u>y = 22.5 L</u>
Answer: There would have to be three nitrogen atoms in the products. The law of conservation of matter states that the amount of substance before a reaction occurs should be the same as the amount of substance after the reaction.
Explanation: This is the EXACT sample answer from the test, just reword it if you want. ^
The number of grams of NaOH that are needed to make 500 ml of 2.5 M NaOH solution
calculate the number of moles =molarity x volume/1000
= 2.5 x 500/1000 = 1.25 moles
mass = moles x molar mass of NaOH
= 1.25 x40= 50 grams of NaOH
<span>The formation of a derivative being a necessary step in the experiment lies in the importance of the derived structure. Often the derived product confers to reaction pathways which uses less reactive starting materials and more easily proceeds to completion. This also allows us to take a small amount of sample. The derived product at times is a general compound allowing its easy analysis. Often we encounter a product but we find it difficult to analyse it in ways we want. Here lies the essence of forming a derivative which often are simpler compounds allowing easier analysis yet having similar functional groups and structural properties. Also sometimes we encounter problems when our desired product is unstable and forms stable degraded products. But if we somehow manage to synthesize a derivative it may be relatively stable and form no degradation products. It would be stable at least for a significant period of time making it easier to study its properties. The derived product also at times are synthesized using general reaction pathways facilitating a way of easier synthesis and helping it to correlate with other similar reaction pathways and products.So the above paragraph accounts for the need of derivatives. When we encounter problems similar to those mentioned above it becomes necessary for a researcher to form rather synthesize a derivative.</span>
