The mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
<h3>What is the mole ratio of a reaction?</h3>
The mole ratio of a reaction is the ratio in which the reactants and products of a given reaction occur for the reaction to proceed to completion.
The mole ratio of a reaction is also known as the stoichiometry of the reaction.
The equation of the two reactions are given below:
From the equation of the reaction reaction, an equal volume of hydrogen gas will be produced by the two reactions.
Therefore, the mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
Learn more about mole ratio at: brainly.com/question/19099163
#SPJ1
Answer:
A. 2NO + O2 -> 2NO2
B. 4Co + 3O2 -> 2Co2O3
C. 2Al + 3Cl2 -> Al2Cl6
D. 2C2H6 + 7O2 -> 4CO2 + 6H2O
E. TiCl4 + 4Na -> Ti + 4NaCl
Answer:
494.1 kPa
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
P1 = 294 kPa
P2 = ?
V1 = 42.9 liters
V2 = 22.8 liters
T1 = 76.0°C = 76 + 273 = 349K
T2 = 38.7°C = 38.7 + 273 = 311.7K
294 × 42.9/349 = P2 × 22.8/311.7
12612.6/349 = 22.8 P2/311.7
36.14 = 22.8P2/311.7
Cross multiply
36.14 × 311.7 = 22.8P2
11264.605 = 22.8P2
P2 = 11264.605 ÷ 22.8
P2 = 494.1 kPa
Answer:
a. Plum pudding model
Explanation:
The plum pudding model of the atom was proposed by J.J. Thomson. It was the model he derived from his experiment on the gas discharge tube.
J.J Thomson was the first person to discover electrons which he called cathode rays because in the discharge tube, they emanate from the cathode.
- This led him to suggest the plum pudding model of the atom.
- The model reflects electrons being surrounded by a volume of negative charges.
Answer:
Sodium Bicarbonate on decomposition produces Carbon dioxide gas and Water vapors.
<span> 2 NaHCO</span>₂<span> </span> →<span> Na</span>₂<span>CO</span>₃<span> (s) </span>+ <span> CO</span>₂<span> (g) + H</span>₂<span>O (g)
</span>
Explanation:
Let suppose you burn 168 g ( 2 moles ) of NaHCO₃, a gas will produced and product is left behind. On measuring the product formed it will be almost equal to 105 g. This shows that the product is Na₂CO₃ and 1 mole of it is being produced after decomposition of sodium bicarbonate.
