Answer is 355 grams.
Explanation:
Given the molecular weights:
M
r
N
a
O
H
=
40
g
m
o
l
M
r
N
a
2
S
O
4
=
142
g
m
o
l
The analogy of the moles will be held constant:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
1
n
N
a
O
H
n
N
a
2
S
O
4
=
2
For each one, substitute:
n
=
m
M
r
Therefore:
n
N
a
O
H
n
N
a
2
S
O
4
=
2
m
N
a
O
H
M
r
N
a
O
H
m
N
a
2
S
O
4
M
r
N
a
2
S
O
4
=
2
200
40
x
142
=
2
200
⋅
142
40
x
=
2
200
⋅
142
=
2
⋅
40
x
x
=
200
⋅
142
2
⋅
40
=
100
⋅
142
40
=
10
⋅
142
4
=
1420
4
=
=
710
2
=
355
g
r
a
m
s
(or just use a calculator)
Answer:
To do this question. we first have to find the mass% of nitrogen in N2O and then, using that percentage, we can simply find the number of moles of N from the number of moles of N2O
<u></u>
<u>Mass % of Nitrogen:</u>
Mass% of nitrogen = (Molar mass of N2 / Molar mass of N2O)*100
Mass% nitrogen = (28 / 44)*100
Mass% of Nitrogen = 0.63 * 100
Mass% nitrogen = 63%
<u>Mass of Nitrogen:</u>
So, now we can say that in any given mass of N2O. 63% of the total mass is the mass of Nitrogen
Hence, total mass * 63/100 = Mass of Nitrogen
Replacing the variables
40 * 0.63 = Mass of Nitrogen
Mass of Nitrogen = 25.2 grams
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
Answer:
a) 7.96* 10⁻³ s
b) 0.0112 s
c) 1.645* 10⁻³⁸ g
Explanation:
for the reaction
Cyclobutane (A) → 2 ethylene (B)
the reaction rate (first order )is
-dCa/dt = k*Ca
∫dCa/Ca = - ∫k*dt
ln(Ca/Ca₀) = -k*t → Ca = Ca₀*e^(-k*t)
therefore
a) the half- life represents the time required for the concentration Ca to drop to half of the initial value ( Ca=Ca₀/2) therefore
Ca₀/2 = Ca₀*e^(-k*t) → - ln 2 = -k*t → t = ln 2 / k =ln 2 / ( 87 1/s) = 7.96* 10⁻³ s
b) Ca = Ca₀*e^(-k*t) , for Ca= Wa/(V*M) , where Wa is weight:
Wa = Wa₀*e^(-k*t)
for Wa₀= 4 g and Wa = 4g - 2.5 g = 1.5 g
→ t= (1/k)* ln(Wa₀/Wa) = 1/( 87 1/s) * ln [ 4g/(1.5 g)] = 0.0112 s
c) for Wa₀= 4 g and t=1 s
Wa = Wa₀*e^(-k*t) = 1 g * e^(- 87 (1/s) *1 s )= 1.645* 10⁻³⁸ g ≈ 0
Answer:
See the answer below, please.
Explanation:
The bonds formed between metals and nonmetals are called ionics. These occur between atoms with electronegativity difference. Example: NaCl (Sodium Chloride)
Instead, covalent bonds are formed between two nonmetals (one or more electron pairs are shared). Example: H202 (hydrogen peroxide).
In the case of metal formed bonds, they are called metallic.
