<u>Given:</u>
Mass of MgBr2 = 0.500 g
<u>To determine:</u>
Number of anions in 0.500 g MgBr2
<u>Explanation:</u>
Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol
Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles
Based on stoichiometry-
1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions
Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-
Now,
1 mole of Br- contains 6.023 * 10²³ anions
0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions
Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2
Answer:
8.13 ×10²³ atoms
Explanation:
Given data:
Mass of magnesium = 32.45 g
Number of atoms = ?
Solution:
Number of moles of Mg:
Number of moles = mass/molar mass
Number of moles = 32.45 g/ 24 g/mol
Number of moles = 1.35 mol
Number of atoms:
1 mole contain 6.022×10²³ atoms
1.35 mol × 6.022×10²³ atoms/ 1mol
8.13 ×10²³ atoms
A base is weak when only a little of it dissolved in Water.
It images help u...
The relation between force, mass of an object and its acceleration is given by following equation
F = mass of object X acceleration
The unit of Force is Newton (N)
unit of acceleration is ms⁻²
unit of mass if Kg
[all these are SI units]
Mass of object =
mass of sprinter =
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.