Answer:
The answer is (e) : phosphoglucomutase, UDP-glucose pyrophosphorylase, glycogen synthase then amylo-(1,4-1,6)-transglycosylase.
Explanation:
Phosphoglucomutase: Convert glucose-6-phosphate to glucose-1-phosphate.
UDP-glucose pyrophosphorylase: Form UDP-glucose from glucose-1-phosphate.
Glycogen synthase: Add the new glucose from UDP-glucose to the growing glycogen chain.
Amylo-(1,4-1,6)-transglycosylase: This is a branching enzyme, it initiates formation of branches evolving from the main chain.
Average atomic mass is the weighted average atomic masses with regard to the relative abundance of the isotopes
average atomic mass of Li = relative abundance of Li-6 x mass of Li-6 + relative abundance of Li-7 x mass of Li-7
average atomic mass of Li = (7.42% x 6.0151 a.m.u) + (92.58% x 7.0160 a.m.u)
= 0.446 + 6.495
= 6.941 amu
average atomic mass of Li is 6.941 amu
Complete Question:
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
Answer:
2.23x10⁶ g
Explanation:
The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:
0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³
0.7976 g/m³
The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:
A = πR², where R is the radius = 1.01x10² m (half of the diameter)
A = π*(1.01x10²)²
A = 32047 m²
The volume is then:
V = 32047 * 87.32
V = 2.7983x10⁶ m³
The mass of the F⁻ is the concentration multiplied by the volume:
m = 0.7976 * 2.7983x10⁶
m = 2.23x10⁶ g
Answer: 1.5 moles
Explanation: one mole Zn uses 2 moles HCl.
1.5 moles Zn uses 3.0 mol HCl. Then Zn is a limiting reactant
And produces equal amount of H2.
<u>Answer:</u> The value of equilibrium constant for the net reaction is 11.37
<u>Explanation:</u>
The given chemical equations follows:
<u>Equation 1:</u> ![A+2B\xrightarrow[]{K_1} 2C](https://tex.z-dn.net/?f=A%2B2B%5Cxrightarrow%5B%5D%7BK_1%7D%202C)
<u>Equation 2:</u> ![2C\xrightarrow[]{K_2} D](https://tex.z-dn.net/?f=2C%5Cxrightarrow%5B%5D%7BK_2%7D%20D)
The net equation follows:
![D\xrightarrow[]{K} A+2B](https://tex.z-dn.net/?f=D%5Cxrightarrow%5B%5D%7BK%7D%20A%2B2B)
As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.
The value of equilibrium constant for net reaction is:
We are given:
Putting values in above equation, we get:
Hence, the value of equilibrium constant for the net reaction is 11.37
