Answer:
c. Many of their bonds are C-C and C-H
Explanation:
The majority of bonds in carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.
These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.
The correct answer is c.
Answer:
I know that the 100-mL graduated cylinders are always read to 1 decimal place.
I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.
Answer:- The hydroxide ion concentration of the solution is 
.
Solution:- The formula used to calculate pOH from hydroxide ion is:
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:
![[OH^-]=10^-^p^O^H](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5Ep%5EO%5EH)
pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:
![[OH^-]=10^-^5^.^7^1](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5E5%5E.%5E7%5E1)
![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
= 0.00000195 or
So, the hydroxide ion concentration of the solution is .
Answer:
16.9g
Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2
Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.
Then to calculate the theoretical yield you need to compare moles of the reactants:
m(Cu)=5g
M(Cu)=63.55
n(Cu)=5/63.55=0.0787
By comparing coefficients you require twice as much silver: 0.157mol
n(Ag)=0.157
M(Ag)=107.86
m(Ag)=0.157x107.86=16.9g
Hence, the theoretical yield of this reaction would be 16.9g
