Answer:
The answer to your question is because of the force of friction between the ball and the ground.
Explanation:
Exactly, Newton's first law states that every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impress on it.
After analyzing Martin's experiment we could conclude that Newton's first law of motion is wrong but we must remember that between a body and the surface there is a force of friction that causes that the object comes to stop.
We can not see this force but it is there.
B. Light energy and heat energy!
This is because the thermal energy which is the heat. As you know fire heats up and burns, a lit candle would do just the seem. Also the fire is bright enough to light up an area!
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
Answer:
The answer to your question is below
Explanation:
I just write the formulas of the reactants and products and balanced the reactions.
a)
3H₂(g) + N₂(g) ⇒ 2NH₃
b)
2K + 2H₂O ⇒ 2KOH + H₂ (g)
c)
2Al(s) + Fe₂O₃ ⇒ Al₂O₃ + 2Fe
d)
Fe₂O₃ + 2Al ⇒ Al₂O₃ + 2Fe
e)
Ba(OH)₂ + 2HBr ⇒ BaBr₂ + 2H₂O
f)
CaCO₃ + Δ ⇒ CaO + CO₂
The grams of CuCl2 required to prepare 1250g of 1.25% by mass is calculated as follow
1250 = 100%
what about 1.25%
by cross multiplication
= 1.25 x1250/100 = 15.63 grams