Answer:
By far the most important use of alkenes is in the making of plastics as plastics are used in almost everything.
Explanation:
Alkenes themselves aren't used much in everyday life however Alkenes are very important to industrial synthesis as it is relatively easy to turn them into other things.
Alkenes can be turned into polymers or plastics through addition reactions and the most common ethene is turned into everything from plastic bags to bottles.
Alkenes can also be turned into alcohols. most commonly propene is used as a feedstock to produce butanol and other products useful in industry or for production
Explanation:
Can you be my friend in here
Answer:
660kcal
Explanation:
The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.
We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:
3.3L * (1000mL / L) * (5g/100mL)= 165 g.
If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.
Answer:
In order of decreasing miscibility
C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)
Explanation:
The solubility of a solid is a measure of its ability to dissolve in a liquid while for liquids, the miscibility is a measure of thhe liquid to mix with anoyjer liquid resulting in a soltion which can hold any amount of either liquids. Immiscible liquids are those that are not soluble or have very limited solibility with each other.
C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)
In the order of decreasing miscibility as like dissolve like, ability to dissociate and polar and organic characteristics are considered
Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>
3.8 /
=
((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
<span>0.925 grams if using hydrochloric acid in the reaction.
0.462 grams if using sulfuric acid in the reaction.
0.000 grams if using nitric acid in the reaction.
Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is:
Zn + 2 HCl ==> ZnCl2 + H2
So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen.
Atomic weight zinc = 65.38
Atomic weight hydrogen = 1.00794
Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol
Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So
0.458855919 * 2 * 1.00794 = 0.92499847 grams.
Rounding to 3 significant figures gives 0.925 grams.
To show the assumption of the acid used, the balanced equation for sulfuric acid would be
Zn2 + H2SO4 ==> Zn(SO4)2 + H2
Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid).
If nitric acid were used, the reaction is
4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O
Which means that NO hydrogen gas is generated.
The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.</span>
