Answer:
There is no picture but the one that is in motion in the picture has kinetic energy.
Explanation:
Kinetic energy is the energy of motion.
hope this helped!
Answer: 65.38g of Ca(OH)2 is needed
Explanation:
From The equation of reaction
2 HCl + Ca ( OH ) 2 ⟶ CaCl 2 + 2 H 2 O
NB: Molar mass of HCl= 1+35.5=36.5
Ca(OH)2= 74
From The stoichiometric equation
2mol of HCl(36.5×2=73) require 1mol of Ca(OH)2 (74g)
Hence 64.5g of HCl will require 64.5×74/73= 65.38g of Ca(OH)2
Answer:
D) 1/2
Explanation:
Using Ideal gas equation for same mole of gas as
Given,
P₂ = 4P₁
T₂ = 2T₁
Using above equation as:

<u>The volume change by half of the original.</u>
Answer:
0.0745 mole of hydrogen gas
Explanation:
Given parameters:
Number of H₂SO₄ = 0.0745 moles
Number of moles of Li = 1.5107 moles
Unknown:
Number of moles of H₂ produced = ?
Solution:
To solve this problem, we have to work from the known specie to the unknown one.
The known specie in this expression is the sulfuric acid, H₂SO₄. We can compare its number of moles with that of the unknown using a balanced chemical equation.
Balanced chemical equation:
2Li + H₂SO₄ → Li₂SO₄ + H₂
From the balanced equation;
Before proceeding, we need to obtain the limiting reagent. This is the reagent whose given proportion is in short supply. It determines the extent of the reaction.
2 mole of Li reacted with 1 mole of H₂SO₄
1.5107 mole of lithium will react with
= 0.7554mole of H₂SO₄
But we were given 0.0745 moles,
This suggests that the limiting reagent is the sulfuric acid because it is in short supply;
since 1 mole of sulfuric acid produced 1 mole of hydrogen gas;
0.0745 mole of sulfuric acid will produce 0.0745 mole of hydrogen gas
Answer:
The percentage efficiency of the electrical element is approximately 82.186%
Explanation:
The given parameters are;
The thermal energy provided by the stove element,
= 3.34 × 10³ J
The amount thermal energy gained by the kettle,
= 5.95 × 10² J
The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

Therefore, we get;

The percentage efficiency of the electrical element, η% ≈ 82.186%.