Answer:
2.5 moles of KOOH are produced.
Explanation:
1)Given data:
Number of moles of KOOH produced = ?
Number of moles of LiClO = 5 mol
Solution:
Chemical equation:
2LiClO + KHSO₄ → Li₂SO₄ + Cl₂ + KOOH
now we will compare the moles of KOOH and LiClO.
LiClO : KOOH
2 : 1
5 : 1/2×5 = 2.5
2.5 moles of KOOH are produced.
<span>a)
</span>First order
in A and zero order in B
<span>ln [A]
= (ln 0.1) (2) + ln Ao = ln 0.01 + ln Ao = ln 0.01 Ao = 1.0% of A will remain</span>
<span>b)
</span>First
order in A and first order in B
<span>1/[A] – 1/[A]0= kt where t+=1 and k=9</span>
[A]/[A]=1/19=0.053=5.3%
<span>c)
</span>Zero
order in both A and B
<span>[A]0-[A] = kt</span>
Then at 2
hours [A]=0 All of it has reacted.
<span> </span>
Since Group 2 alkali earth metals have 2 valence electrons, they tend to lose those 2 when forming ionic bonds. And the Loss of Electrons = Oxidation (L.E.O. for short). Therefore this group, including Mg and Ca, have an oxidation of [+2].
So the correct answer is C) +2
The runner ran a total of 8800 yards.
Answer:
V = 12.93 L
Explanation:
Given data:
Number of moles = 0.785 mol
Pressure of balloon = 1.5 atm
Temperature = 301 K
Volume of balloon = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values.
V = nRT/P
V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm
V = 19.4 L /1.5
V = 12.93 L
