Answer:
Option (D)
Explanation:
Phosphorylation can be simply defined as the addition of a phosphate group to an organic and inorganic molecule. This process helps in regulating the processes that occur in the cells. It leads to the growth and development of cells and this process is efficiently carried out with the help of enzymes like kinase. It also plays an important role in transferring the signals within the cells, synthesis, and functioning of proteins within the cells, and storing as well as releasing of energy.
Thus, the correct answer is option (D).
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
3.0 × 10¹¹ RBC's (or) 3E11 RBC's
Solution:
Step 1: Convert mm³ into L;
As,
1 mm³ = 1.0 × 10⁻⁶ Liters
So,
0.1 mm³ = X Liters
Solving for X,
X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³
X = 1.0 × 10⁻⁷ Liters
Step 2: Calculate No. of RBC's in 5 Liter Blood:
As given
1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's
So,
5.0 Liters of Blood will contain = X RBC's
Solving for X,
X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters
X = 3.0 × 10¹¹ RBC's
Or,
X = 3E11 RBC's
Explanation:
sodium atom is neutral while sodium ion is a charged specie with a charge of +1
Answer:
1.772 gram is the approximate answer
Explanation:
molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole
the reaction is
AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl
from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3
implies 132g AlCl3 gives 78g Al(OH)3
Implies 3g AlCl3 gives
3*122/78 = 1.772 grams
