Answer:
The empirical formula of the compound is =
The name of the compound is potassium bromate.
Explanation:
Mass of potassium = 4.628 g
Moles of potassium =
Mass of bromine = 9.457 g
Moles of bromine =
Mass of oxygen = 5.681 g
Moles of oxygen =
For empirical formula of the compound, divide the least number of moles from all the moles of elements present in the compound:
Potassium :
Bromine;
Oxygen ;
The empirical formula of the compound is =
The name of the compound is potassium bromate.
Answer:
1.728 mol /(L*min)
Explanation:
Hello,
In the attached photo, you'll find the numerical procedure for your question.
- Take into account that the negative sign is eligible for reagents and positive for products.
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Answer:
20 molecules of oxygen gas remains after the reaction.
Explanation:
Molecules of ethyne = 52
Molecules of oxygen gas = 150
According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.
Then 52 molecules of ethyne will react with:
of oxygen gas.
As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.
Remaining molecules of recessive reagent = 150 - 130 = 20
20 molecules of oxygen gas remains after the reaction.
Answer:
The answer to your question is V2 = 1.82 l
Explanation:
Data
Volume 1 = 77 l
Pressure 1 = 18 mmHg
Volume 2 = ?
Pressure 2 = 760 mmHg
Process
Use Boyle's law to solve this problem
P1V1 = P2V2
-Solve for V2
V2 = P1V1/P2
-Substitution
V2 = (18 x 77) / 760
-Simplification
V2 = 1386 / 760
-Result
V2 = 1.82 l
Answer:
The water will evaporate and fly out of the bucket; the process will not stop until there is enough water vapor in the atmosphere that the vapor pressure stops the water from boiling further.
Explanation: