441 g CaCO₃ would have to be decomposed to produce 247 g of CaO
<h3>Further explanation</h3>
Reaction
Decomposition of CaCO₃
CaCO₃ ⇒ CaO + CO₂
mass CaO = 247 g
mol of CaO(MW=56 g/mol) :
From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :
mass CaCO₃(MW=100 g/mol) :
Answer:
Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao
Explanation:
To find this, we will use this formula:
Molar mass of element
------------------------------------ x 100
Molar mass of compound
So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.
Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol
When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!
95.994
-----------
164.086
When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.
So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.
Although you have not provided the circled electron, I can help you with a wide explanation.
1) Atomic number of manganese is 25. That means that it has 25 protons and 25 electrons.
2) Those 25 electrons are distributed (electron configuration) as per the quantum rules:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
3) The most reasonable is that you have been asked to give the possible quantum numbers for an electron in the 4s or 3d.
4) Those are 7 electrons and these are their possible sets of quantum numbers:
i) For the two electrons in 4s:
n is the main energy level so n = 4
l tells the kind of orbital, which is s, so l = 0
ml is also 0 (it can be from -l to + l, so given that l i s0, ml is 0)
ms: one is +/12 and the other is -1/2 (this is the spin number).
ii) For the 5 electrons in 3d
n = 3
l can be 0, 1, or 2
if l = 0, then ml = 0
if l = 1, then ml can be -1, 0 , or 1 (from - l to + l)
ms can be either +1/2 or - 1/2 (spin)
Answer:1.
1.Balanced equation
C4H10 + 9 02 ==> 5H20 +4CO2
2. Volume of CO2 =596L
Explanation:
1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;
CxHy +( x+y/4) O2 ==> y/2 02 + xCO2
Where x and y are number of carbon and hydrogen atoms respectively.
For butane (C4H10)
x=4 and y=10
Therefore
C4H10 + 9 02 ==> 5H20 +4CO2
2. Mass of butane = 0.360kg
Molar mass of C4H10 = ( 12×4 + 1×10)
= 48 +10=58g/mol= 0.058kg/mol
Mole = mass/molar mass
Mole = 0.360/0.058= 6.2moles
From the stoichiometric equation
1mole of C4H10 will gives 4moles of CO2
Therefore
6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2
Using the ideal gas equation
PV=nRT
P= 1.0atm
V=?
n= 24.8mol.
R=0.08206atmL/molK
T=20+273=293
V= 24.8 × 0.08206 × 293
V= 596L
Therefore the volume of CO2 produced is 596L
