Net overall dissociation:
Na2CO3 ---> 2Na(-) + CO3(2-)
*The ion charge is in parenthesis
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Answer:
![\boxed{\text{300 g}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B300%20g%7D%7D)
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32 60
CH₃OH + CO ⟶ CH₃COOH
m/g: 160
(a) Moles of CH₃OH
![\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20CH%24_%7B3%7D%24OH%7D%20%3D%20%5Ctext%7B160%20g%20CH%24_%7B3%7D%24OH%20%7D%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20CH%24_%7B3%7D%24OH%20%7D%7D%7B%5Ctext%7B32%20g%20CH%24_%7B3%7D%24OH%7D%7D%3D%20%5Ctext%7B5.00%20mol%20CH%24_%7B3%7D%24OH%7D)
(b) Moles of CH₃COOH
![\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20CH%24_%7B3%7D%24COOH%7D%20%3D%20%5Ctext%7B5.00%20mol%20CH%24_%7B3%7D%24OH%20%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20CH%24_%7B3%7D%24COOH%7D%7D%7B%5Ctext%7B1%20mol%20CH%24_%7B3%7D%24OH%20%7D%7D%20%3D%20%5Ctext%7B5.00%20mol%20CH%24_%7B3%7D%24COOH%7D)
(c) Mass of CH₃COOH
![\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20CH%24_%7B3%7D%24COOH%7D%20%3D%5Ctext%7B5.00%20mol%20CH%24_%7B3%7D%24COOH%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B60%20g%20CH%24_%7B3%7D%24COOH%7D%7D%7B%5Ctext%7B1%20mol%20CH%24_%7B3%7D%24COOH%7D%7D%20%3D%20%5Ctextbf%7B300%20g%20CH%24_%7B3%7D%24COOH%7D%5C%5C%5C%5C%5Ctext%7BThe%20maximum%20mass%20of%20ethanoic%20acid%20that%20can%20be%20produced%20is%20%24%5Cboxed%7B%5Ctextbf%7B300%20g%7D%7D%24%7D)
Answer :
The steps involved in the electron dot structure of
are :
First we have to determine the total number of valence electron in
.
Number of valence electrons in N = 5
The charge on N is (-3). So, we add 3 electrons.
Total number of valence electrons = 5 + 3 = 8 electrons
The image is shown below.
The expected speed is v = 85.5 km/h
v = 85.5 km/h = (85.5 km/h)*(0.2778 (m/s)/(km/h)) = 23.75 m/s
If there is an uncertainty of 2 meters in measuring the position, then within a 1-second time interval:
The lower measurement for the speed is v₁ = 21.75 m/s,
The upper measurement for the speed is v₂ = 25.75 m/s.
The range of variation is
Δv = v₂ - v₁ = 4 m/s
The uncertainty in measuring the speed is
Δv/v = 4/23.75 = 0.1684 = 16.84%
Answer: 16.8%