Explanation:
A catalyst lowers the activation energy of a reaction allowing them to proceed faster than they would naturally. Activation energy is the free energy that is required to be input in the reactant side to activate them to the transition state after which the reaction proceeds spontaneously to products.
An example of a catalyst is platinum, that is put in the exhaust of cars, to help convert carbon monoxide to carbon dioxide before it is emitted into the air.
Answer: they lower these activation required for the reactions to take place.
Explanation:
Answer
its B
Explanation:
B.
Nuclear energy produces little air pollution.
Is mass conserved when 10 g of sodium hydroxide undergoes a chemical change during an interaction with 57 g of copper sulfate? Use complete sentences to support your answer by explaining how this can be demonstrated.
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please use complete sentences for the answer/an explanation, i'm confused on this questionn v( ‘.’ )v ...
Answer:Instead of “theory” scientists use the word hypothesis to describe a possible explanation for a scientific mystery. Theories in science are hypotheses that.
Explanation:
what is the reacted volume of oxygen gas at STP and the mass of the produced water vapour in this reaction ?
Answer:
Volume of O₂ = 56 dm³
mass of water vapors (H₂O) = 90 g
Explanation:
Data Given:
mass of Oxygen = 10 g
Volume of Oxygen = ?
mass of the water vapor = ?
Reaction Given:
2H₂+O₂---->2H₂O
Solution:
First we have to look at the reaction for the information required
2H₂ + O₂ -------> 2H₂O
2 mol 1mol 2 mol
now convert moles to grams
molar mass of H₂ = 2(1) = 2 g/mol
molar mass of O₂ = 2(16) = 32 g/mol
molar mass of H₂0 = 2(1) + 16 = 18 g/mol
So the masses will be
2H₂ + O₂ -------> 2H₂O
2 mol (2 g/mol) 1mol (32 g/mol) 2 mol (18 g/mol)
4 g 32 g 36 g
So now we know that
4 g of hydrogen combine with 32 g of Oxygen and give 36 g of water vapors.
By using above information
First we find the volume of Oxygen:
For this first we find mass and then moles of Oxygen
As we know
if 4 g of hydrogen combine with 32 g of Oxygen then how much oxygen will react with 10 g of hydrogen
Apply unity formula
4 g of hydrogen H₂ ≅ 32 g of Oxygen O₂
10 g of hydrogen H₂ ≅ X g of Oxygen O₂
by doing Cross multiplication
g of Oxygen O₂ = 32 g x 10 g / 4 g
g of Oxygen O₂ = 80 g
So,
mass of oxygen = 80 g
now find moles of oxygen
formula used:
no. of moles = mass in grams/ molar mass . . . . . . (1)
Put values in above equation 1
no. of moles = 80 g / 32 g/mol
no. of moles = 2.5
Now to find volume of oxygen
Formula used
Volume of O₂ = no. of moles x molar volume (22.4 dm³/ mol) . . . . . . (2)
Put values in equation 2
Volume of O₂ = 2.5 moles x 22.4 dm³/mol
Volume of O₂ = 56 dm³
______________________
Now to find mass of water vapors
As we now
if 4 g of hydrogen produce 36 g of water vapors then how much water vapor will produce from 10 g of hydrogen
Apply unity formula
4 g of hydrogen H₂ ≅ 36 g of water vapors (H₂O)
10 g of hydrogen H₂ ≅ X g of water vapors (H₂O)
by doing Cross multiplication
g of water vapors (H₂O) = 36 g x 10 g / 4 g
g of water vapors (H₂O) = 90 g
So,
mass of water vapors (H₂O) = 90 g