3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and the last O atoms
Reaction
Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O
Zn : left = a, right =1 ⇒a=1
H : left = b, right = 2d⇒ b=2d (eq 1)
N : left = b, right = 2+c⇒b=2+c (eq 2)
O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)
3(2d)=6+c+d
6d=6+c+d
5d=6+c (eq 4)
3(2+c)=6+c+d
6+3c=6+c+d
2c=d (eq 5)
5(2c)=6+c
10c=6+c
9c=6
c = 2/3
d = 2 x 2/3
d = 4/3
b = 2 x 4/3
b = 8/3
The equation
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to
Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3
3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
Answer:
a mixture in which the composition is uniform throughout the mixture.
Explanation:
Answer: The kilograms of water must evaporate from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
Explanation:
According to the ratio and proportion:
where,
= concentration of ist solution = 25%
= mass of ist solution = 8 kg
= concentration of second solution = 40%
= mass of second solution = ? kg
Thus the final solution must have a mass of 5 kg , i.e (8-5)= 3 kg of mass must be evaporated.
Therefore, the mass that must be evaporated from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
