Answer: 1.07×10^-20microlitre
Explanation:
1cm3 = 1000microlitres
1.07×10^-23 cm3 of tungsten = 1.07×10^-23 x 1000 = 1.07×10^-20microlitre
The empirical formula is C₇H₆O₂.
Assume that you have 100 g of the compound.
Then you have 68.84 g C and 4.962 g H.
Mass of O = (100 – 68.84 – 4.962) g = 26.20 g O.
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
<u>Element</u> <u>Mass/g</u> <u>Moles</u> <u>Ratio</u> <u> ×2</u> <u>Integers</u>
C 68.84 5.732 3.501 7.001 7
H 4.962 4.923 3.006 6.012 6
O 26.20 1.638 1 2 2
The empirical formula is C₇H₆O₂.
Answer:
x = 2.37
Explanation:
ln x = 0.863
apply e to both sides of the equation:
e^(ln x) = e^0.863
e^(ln x) = x = e^0.863
{Demonstration of e^(ln x) = x:
y = e^(ln x)
apply ln to both sides of the equation:
ln y = ln (e^(ln x))
apply logarithmic property (ln xⁿ = n ln x)
ln y = ln x * ln e (ln e = 1)
ln y = ln x
Then:
y = x
Because y = e^(ln x) and x = y, then:
x = e^(ln x)
End of the demonstration }
Returning to our equation:
e^(ln x) = e^0.863
Then,
x = e^0.863 = 2.37
Answer:
2KClO3 ==> 2 KCl + 3O2
393g O2 x 1 mole/32 g = 12.28 moles O2
moles KClO3 needed = 12.28 mol O2 x 2 mol KClO3/3mol O2 = 8.19 mol
grams KClO3 = 8.19 moles x 123g/mol = 1007g
Explanation:
Answer:
Following are the chemical equation to the given question:
Explanation:
The Electrode is a silver film that is covered with such a thin coating of silver chloride, either by dipping its wire directly into silver-molten chloride, plating the wire using hydrogen peroxide, or oxidation silver in a chloride. In the given silver electrode, this anode acts as a cathode and thus reduces.
Half of the response reduction:
Half-effect oxidation:
Complete reaction: