Answer:
One way to find out is to change “passive mass” - the dirt and rocks of Earth - and convert it into “active mass” - add energy “E” to it. If we increase the energy “E” we will increase its mass. The key is, how do you turn “passive mass” into “active mass” by adding energy “E” to the “passive mass”? After all, isn’t “passive mass” of the Earth just the dirt and rocks that make up the planet?
Explanation:
Answer:
1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
2) The amount (in grams) of excess reactant H₂ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
<em>N₂ + 3H₂ → 2NH₃.</em>
<em>1) To determine the limiting reactant of the reaction:</em>
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
- We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>
The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).
The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>
- As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
- Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
- The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
Answer:
This reaction has infinite ways to be balanced
Explanation:
To balance this equation we can use the algebraic method:
N2O4(g) + CO → NO(g) + CO2(g)+NO2(g)
Where we write each molecule as a letter:
A + B → C + D + E
Then, we write the equations according the number of atoms of each molecule. That is:
Oxygen → 4A + B = C + 2D + 2E <em>(1)</em>
Nitrogen → 2A = C + E <em>(2)</em>
Carbon → B = D <em>(3)</em>
Then, we have to give 1 arbitral number for a letter. For example:
B = 1; D = 1
<em>(1) </em>4A + 1 = C + 2 + 2E
4A = C + 2E + 1
2A = C + E <em>(2) </em>Twice <em>(2):</em>
4A = 2C + 2E
Subtracting <em>(1) </em>in <em>(2)</em>
C + 2E + 1 = 2C + 2E
C + 1 = 2C
1 = C
Si 1 = C:
4A + 1 = 1 + 2 + 2E
4A = 2 + 2E <em>(1)</em>
y:
2A = 1 + E <em>(2)</em>
Twice:
4A = 2 + 2E
As <em>(1) </em>and <em>(2) </em>are the same equation:
<h3>This reaction has infinite ways to be balanced</h3><h3 />
For example:
N2O4(g) + CO → NO(g) + CO2(g)+NO2(g)
1. CHEM CHANGE
2. PHYS CHANGE
3. CHEM CHANGE
4. CHEM CHANGE
5. PHYS CHANGE
To solve this question,
let us first calculate how much all the nucleons will weigh when they are apart,
that is:
<span>Mass of 25 protons = 25(1.0073) = 25.1825 amu </span>
Mass of neutrons = (55-25)(1.0087) = 30.261 amu
So, total mass of nucleons = 30.261+25.1825 =
55.4435 amu
<span>Now we subtract the mass of nucleons and mass of the Mn
nucleus:
55.4435 - 54.938 = 0.5055 amu
This difference in mass is what we call as the mass defect of
a nucleus. Now we calculate the binding energy using the formula:</span>
<span> E=mc^2 </span>
<span>But first convert mass defect in units of SI (kg):
Δm = 0.5055 amu = (0.5055) / (6.022x10^26)
<span>Δm = 8.3942x10^-28 kg</span>
Now applying the formula,
E=Δm c^2
E=(8.3942x10^-28)(3x10^8)^2
E=7.55x10^-11 J</span>
Convert energy from Joules
to mev then divide by total number of nucleons (55):
E = 7.55x10^-11 J *
(6.242x10^12 mev / 1 J) / 55 nucleons
<span>E = 8.57 mev / nucleon</span>
