Answer:
310.53 g of Cu.
Explanation:
The balanced equation for the reaction is given below:
CuSO₄ + Zn —> ZnSO₄ + Cu
Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:
Molar mass of CuSO₄ = 63.5 + 32 + (16×4)
= 63.5 + 32 + 64
= 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g
Summary:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.
Thus, 310.53 g of Cu were obtained from the reaction.
The correct option is A.
To calculate the binding energy, you have to find the mass defect first.
Mass defect = [mass of proton and neutron] - Mass of the nucleus
The molar mass of thorium that we are given in the question is 234, the atomic number of thorium is 90, that means the number of neutrons in thorium is
234 - 90 = 144.
The of proton in thourium is 90, same as the atomic number.
Mass defect = {[90 * 1.00728] +[144* 1.00867]} - 234
Note that each proton has a mass of 1.00728 amu and each neutron has the mass of 1.00867 amu.
Mass defect = [90.6552 + 145.24848] - 234 = 1.90368 amu.
Note that the unit of the mass is in amu, it has to be converted to kg
To calculate the mass in kg
Mass [kg] = 1.90368 * [1kg/6.02214 * 10^-26 = 3.161135 * 10^-27
To calculate the binding energy
E = MC^2
C = Speed of light constant = 2.9979245 *10^8 m/s2
E = [3.161135 * 10^-27] * [2.9979245 *10^8]^2
E = 2.84108682069 * 10^-10.
Note that we arrive at this answer because of the number of significant figures that we used.
So, from the option given, Option A is the nearest to the calculated value and is our answer for this problem.
FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :

Answer:
See explanation.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to firstly recall the electron configuration of hydrogen:

To realize that the principal quantum number is 1, the angular is 0 as well as the magnetic one; therefore we infer that all the given n's are not allowed, just l=0 is allowed as well as ml=0 yet the rest, are not allowed.
Best regards!
Answer:
Explanation:
Using Gay lussac's law equation as follows;
P1/T1 = P2/T2