Balanced chemical equation:
2 H2 + 1 O2 = 2 H2O
4 g H2 -------> 32 g O2 -----------> 36 g H2O
↓ ↓ ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?
32 * mass H2O = 2.0 * 36
32 * mass H2O = 72
mass of H2O = 72 / 32
mass of H2O = 2.25 g
hope this helps!.
Answer:
It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.
Explanation:
The question is not very much clear.
If you are asking for molecules then 1 mole water= 6.023 * 10^23
If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3
If you are asking for particles then,
So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.
I hope that was helpful!
H=1 proton,1 electron
O=8 protons,8 neutrons and 8 electrons
total particles in one H2O molecule-28
total no. of particles in 1 mole of water- 6.023 * 10^23 * 28
When sodium amide i.e.
reacts with water i.e. 
results in the formation of sodium hydroxide i.e. 
and ammonia 
.
The chemical reaction is given by:
Now, when ammonia i.e. reacts with water results in the formation of ammonium hydroxide i.e. 
The chemical reaction is given by:
Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).
The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.
Chemical<span> reactions takes place in plants and animals, this result in the formation of substances in some plants and animals that can be used to treat illness. </span>Chemistry<span>is </span>important<span> to everyday </span>life<span>, because it provides medicine. The food we consume each day comes directly from </span>chemical<span> processes.</span>
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
