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Answer:</h3>
8CO₂
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Explanation:</h3>
We are given;
- Butane is a hydrocarbon in the homologous series known as alkane.
We are required to determine the other product produced in the combustion of butane apart from water.
- We know that the complete combustion of alkane yields carbon dioxide and water.
- Therefore, combustion of butane will yield carbon dioxide and water.
- The balanced equation for the complete combustion of butane will be;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Answer:
A.
Explanation:
You should NEVER eat or drink anything in a lab area. You never know what chemicals or gases are in the lab, and they can harm you.
Wearing a drawstring hoodie won't protect you from chemicals.
Don't wait to clean up chemicals, immediately get a teacher and clean it up (follow the teachers instructions). You never know what has spilled, and if it is harmful or not, or if there is a certain procedure to clean it up.
Don't change the equipment in the middle of an experiment. This can tamper with your results, and depending on what you are working with, this can be dangerous.
At the beginning of the war the Northern states had a combined population of 22 million people. The Southern states had a combined population of about 9 million. This disparity was reflected in the size of the armies in the field. The Union forces outnumbered the Confederates roughly two to one.
Also this should be listed for history
Answer: 30.978
Explanation:
From the equation 2 moles of Fe will result in 3 moles copper
so .325 moles Fe will result in .4875 moles Cu
Cu weights 63.546 gm per mole
.4875 moles * 63.546 gm / mole = 30.978 gm of Cu
You need the set of reactions that goes from ammonia to nitric acid.
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1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)-->2NO2(g)
3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)
State the ratio of moles of HNO3 to NH3:
4 moles of NH3 produce 4 mole of NO,
4 moles of NO produce 4 moles of NO2
4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.
=> (8/3) moles HNO3 : 4 moles NH3
Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution
M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3
Use proportions:
(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x
=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3
Convert moles to grams:
molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol
mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g
Answer: 3213 g.
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