When q is the heat energy in joules (J)
so, according to this formula, we can get q (in joule unit):
q = M*C*ΔT
when M is the mass of the water sample = 1.85 g
C is the specific heat capacity of water = 4.18 J/g.°C
and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C
So, by substitution, we will get the value of q ( in Joule):
∴ q = 1.85 g * 4.18 J/g.°C * 11 °C
= 85 J
Answer:
9 L
Explanation:
According to the question , the given reaction is -
2NO(g) + O₂(g)------->2NO₂(g)
Since ,
At STP ,
One mole of a gas occupies the volume of 22.4 L.
Hence , as given in the question -
9 L of NO , i.e .
22.4 L = 1 mol
1 L = 1 / 22.4 mol
9 L = 1 / 22.4 * 9 L = 0.40 mol
From the chemical reaction ,
The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .
Hence ,
2 moles of NO will produce 2 moles of NO₂.
Therefore ,
0.40 mol of NO will produce 0.40 mol of NO₂.
Hence , the volume of NO₂ can be calculated as -
1 mol = 22.4 L
0.40 mol = 0.40 * 22.4 L = 9 L
Answer:
O2 is limiting reactant
Explanation:
To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:
<em>Moles C6H12O6:</em>
650g * (1mol/180.16g) = 3.608 moles C6H12O6
<em>Moles O2:</em>
650g * (1mol/32g) = 20.31 moles O2
Now, for a complete reaction of 3.608 moles of C6H12O6 are required:
3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2
As there are just 20.31 moles of O2,
<h3>O2 is limiting reactant</h3>
Yes, you're right the answer is 0,02 moles.
Answer:
Sunlight
Water
And carbon dioxide
Additional information :
6CO2 + 6H2O → C6H12O6 + 6O2
