Answer:
A mixture of 2.00 moles of H., 3.00 moles of NH3, 4.00 moles of Co, and 5.00 moles.
Explanation:
A mixture of 2.00 moles of H., 3.00 moles of NH3, 4.00 moles of Co, and 5.00 moles.
The oxidation state of potassium ion K = +1
The oxidation state of oxygen ion O = -2
So, the oxidation state of O2 is = -2 x 2 = -4
Since, KBrO2 is neutral so,
(+1) + (x) + (-4) = Zero
-3 + X = Zero
So, X = +3
The oxidation state of individual bromine atom in KBrO2 is +3
Answer:
Option D is correct.
H₂O + CO₂ → H₂CO₃
Explanation:
First of all we will get to know what law of conservation of mass states.
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Example:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.
For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
Mass percentage composition of carbon in the compound is the composition of carbon in the compound. This can be calculated by finding out the mass of carbon in the compound and the mass of the whole compound .
The compound formula is C2H5Cl
Mass of 1 mol of compound -64.5 g
Mass of carbon - 24 g
Mass percentage = 24/64.5 x100
Therefore mass composition percentage of carbon = 37.2%
