What is the oxidation number of Chlorine in NaClO4?
2 answers:
<u>Answer:</u> Oxidation state of chlorine in the given compound is +7.
<u>Explanation:</u>
Oxidation state is defined as the number which is assigned to the element when it gains or loose electrons. If the element gains electron, it will attain a negative oxidation state and if the element looses electrons, it will attain a positive oxidation state.
We take the oxidation state of chlorine atom be 'x'.
Oxidation state of sodium atom = +1
Oxidation state of oxygen atom = -2
Evaluating the oxidation state of chlorine atom:
Hence, the oxidation state of chlorine in the given compound is +7.
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Answer:
The theoretical yield of urea = <u>120.35kg</u>
The percent yield for the reaction = <u>72.70%</u>
Explanation:
Lets calculate -
The given reaction is -
→
Molar mass of urea = 60g/mole
Moles of =
(since
)
= 4011.76 moles
Moles of =
=
= 2386.36 moles
Theoritically , moles of required = double the moles of
but , , the limiting reagent is
Theoritical moles of urea obtained =
=
Mass of 2005.88 mole of =
= 120352.8g
= 120.35kg
Therefore , theroritical yeild of urea = 120.35kg
Now , Percent yeild =
72.70%
Thus , the percent yeild for the reaction is 72.70%
Answer:
The first thing we have to do is change and state all the units so that we can use our ideal gas law equation ().
650 mmHg is a pressure unit, we have to convert this to kiloPascals. We know that 760 mmHg gives us 101 kPa.
P = 86kPa
T = 15°C + 273K = 288K
R (Gas constant) = 8.31 kj/mol
Molar mass of Ammonia () = (1 x 3) + (14) = 17g/mol
n (moles) =
3.34 mol
V = ?
Rearrange the equation to solve for Volume:
Substitute the values inside:
V =
<u>Therefore 93 L of volume is occupied by the ammonia gas.</u>
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