Answer:
41.63g
Explanation:
Given parameters:
Volume of CaCl₂ = 500mL = 0.5L
Concentration = 0.75mol/L
Unknown:
Mass of the solute needed = ?
Solution:
The mass of the solute can be derived using the expression below;
Mass = number of moles x molar mass
But,
Number of moles = Concentration x Volume
So;
Mass = Concentration x Volume x molar mas
Molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol
Mass = 0.75 x 0.5 x 111 = 41.63g
Answer;
4.5 m³
Solution:
The statement says that two blocks are present on a lid of a container with volume of 9 m³. The mass of lid is equal to the mass of two blocks. It means that initially there are four blocks (or four atm pressure) upon 9 m³ volume.
After that four more blocks are placed on the lid. Means the pressure is increased from 4 atm to 8 atm (2 atm of lid, 2 atm of old blocks, 4 atm of new four blocks).
So, Data generated is,
P₁ = 4 atm
V₁ = 9 m³
P₂ = 8 atm
V₂ = ?
According to Boyle's Law,
P₁ V₁ = P₂ V₂
Solving for V₂,
V₂ = P₁ V₁ / P₂
Putting values,
V₂ = (4 atm × 9 m³) ÷ 8 atm
V₂ = 4.5 m³
Boron: isotope data. Both isotopes ofBoron, B-10 and B-11, are used extensively in the nuclear industry. B-10 is used in the form of boric acid as a chemical shim in pressurized water reactors while in the form of sodium pentaborate it is used for standby liquid control systems in boiling water reactors
Answer:
Option B: It remains unused during the reaction
Explanation:
Just took the test and got it right.
