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>
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
</span>
The answer is A because it has the most amount of speed, the plane has the same speed but doesn't tell you for what amount of time is the speed so what if the plane traveled 15 per 2 hours or 3 obviously the truck is going faster and it if said 15 miles per minute then it would obviously be the plane but since it doesn't specify then the truck would be you rmost likely answer
Hope this helps
Answer:
b) Both p orbitals are perpendicular to the F−Be−F bond axes.
Explanation:
Be has 2 electrons in its valence shell, subshell s is fulfilled, so it has no unpaired electrons in its ground state to make bonds with F. So, it can promote the electrons to the 2p orbital and will having sp hybridization.
The bond between the orbitals sp and the p orbital of F are in opposite directions but the same ax. The two bonds are equivalent, and the molecule had a linear geometry. The two unhybridized p orbitals on Be are vacant, and so they are perpendicular to the F-Be-F bond axes.
Answer:
131 K
Explanation:
The only variables are volume and temperature, so we can use Charles' Law:
V1/T1 = V2/T2
Data:
V1 = 2.00 L; T1 = 210 K
V2 = 1.25 L; T2 = ?
Calculation:
2.00/210 = 1.25/T2
Multiply each side by the lowest common denominator (210T2)
2.00T2 = 1.25 ×210
2.00T2 = 262.5
T2 = 262.5/2.00 = 131 K
The gas was cooled to 131 K.
