Answers
Answer:
Moles of hydrogen formed = 3.5 moles
Explanation:
Given that:-
Moles of aluminium= 4.0 mol
Moles of hydrogen bromide = 7.0 mol
According to the reaction:-
2 moles of aluminum react with 6 moles of hydrogen bromide
1 mole of aluminum react with 6/2 moles of hydrogen bromide
4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide
Moles of hydrogen bromide = 12 moles
Available moles of hydrogen bromide = 7.0 moles
Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)
The formation of the product is governed by the limiting reagent. So,
6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen
1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen
7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen
Moles of hydrogen formed = 3.5 moles
Answer:
3.5 mol H2, HBr (limiting reactant)
Explanation:
4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2
7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2
Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.
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Write a balanced nuclear equation for the following: The nuclide uranium-238 undergoes alpha emission.
O As potential energy increases, kinetic energy decreases.
As potential energy decreases, kinetic energy decreases.
O Potential and kinetic energy are two separate things and have no relationship.
Answers
Answer:
As potential energy increases, kinetic energy decreases
It's an inverse relationship
Final answer:
Potential energy and kinetic energy are related through the law of conservation of energy. When potential energy increases, kinetic energy decreases, and vice versa, as energy is simply converted between these two forms.
Explanation:
The relationship between kinetic energy and potential energy lies within the law of conservation of energy. This law states that energy cannot be created or destroyed, only converted from one form to another.
Potential energy is the energy stored in an object due to its position in a force field or due to its configuration. For example, when you lift a book off the ground, it gains potential energy because of the work done against the force of gravity.
On the other hand, kinetic energy is the energy of an object due to its motion. The same book, when dropped, loses potential energy and gains kinetic energy as it falls towards the ground.
Hence, when potential energy increases, kinetic energy decreases, and vice versa. This is because the total energy (potential + kinetic) must be conserved, assuming no energy is lost to other forms like heat or sound.
Learn more about Conservation of Energy here:
#SPJ12
Answers
Answer:
Because the water is filled up with the sand every where
Explanation:
So the exess sand goes to the bottem
Answer:
Mass/Volume
Explanation:
The sand eventually makes it way to the bottom because of its mass/volume compared to the waters density. Just like while swimming in a pool, we sink to the bottom because of our mass/volume.
Answers
Answer:
3,8×10⁻⁵ mol/L of potassium permanganate solution
Explanation:
To calculate concentration in mol/L you must convert the 3,8 umol to moles and 100 mL to liters, knowing 1 umol are 1×10⁻⁶mol and 1L are 1000 mL.
3,8 umol × (1×10⁻⁶mol / 1 umol ) = 3,8×10⁻⁶mol of potassium permanganate.
100 mL × ( 1L / 1000 mL) = 0,100 L
Thus, concentration in mol/L is:
3,8×10⁻⁶mol / 0,100 L = 3,8×10⁻⁵ mol/L of potassium permanganate solution
I hope it helps!
Answers
Answer:
even I have the same dought
Answers
Answer:
.
Explanation:
Answers
Answer:
Cd²⁺(aq) + 2 e⁻ → Cd(s)
Mg(s) → Mg²⁺(aq) + 2 e⁻
Cd²⁺(aq) + Mg(s) → Cd(s) + Mg²⁺(aq)
Explanation:
A voltaic cell is composed of two half-reactions:
Reduction (cathode): Cd²⁺(aq) + 2 e⁻ → Cd(s)
Oxidation (anode): Mg(s) → Mg²⁺(aq) + 2 e⁻
Cd²⁺ reduces to Cd (the oxidation number decreases from 2 to 0), whereas Mg oxidizes to Mg²⁺ (the oxidation number increases from 0 to 2). We can get the overall cell reaction by adding both half-reactions:
Cd²⁺(aq) + 2 e⁻ + Mg(s) → Cd(s) + Mg²⁺(aq) + 2 e⁻
Cd²⁺(aq) + Mg(s) → Cd(s) + Mg²⁺(aq)