Answer : The mass of carbon required is, 309.4 grams.
Explanation : Given,
Mass of oxygen = 1100 g
Molar mass of oxygen = 16 g/mol
Molar mass of carbon = 12 g/mol
First we have to calculate the moles of oxygen.
Now we have to calculate the moles of carbon.
Moles of carbon =
Moles of carbon =
Now we have to calculate the mass of carbon.
Thus, the mass of carbon required is, 309.4 grams.
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Chromium is a metal in nature. So when one chromium is
bonded to another chromium, there is a weak intermolecular forces which helds
them together which we call as “metallic bonding”.
Metallic bonding is the intermolecular force of attraction which
exist between valence electrons and the metal atoms. It is considered as the
sharing of various detached electrons between many positive ions, whereby the
electrons serve as a "glue" which gives the substance a definite
structure.
I got 9.25527 for the pKa
<span>then for the Henderson-Hasselbalch equation: </span>
<span>9.00=9.25527+log(.600/acid) </span>
10^(-.25527)=(.600/acid)correct to this line. Then
0.5556 = 0.6/acid and
acid = 0.6/0.5556 = 1.08 which is the reciprocal of your number)
<span>.5556/.600M=acid </span>
<span>acid=.925925...M </span>
<span>(.925925)*2.10L= Molarity</span>