Carbon dioxide is a carbon atom and two oxygen atoms, therefore CO2
For the others, they are hydrocarbons.
The first part of the name is determined by how many carbon atoms there are. The second part, is by the type. Alcohol, Alkane, Alkene, alkynes, acid, esters, amides.
Answer:
Let's say we were Subtracting 3-2=? To Find the Answer we would Subtract 2 from 3 which is 1 Simple our answer is 1 But let's say the Question is 3 - 1= ? to find this answer we would subtract 1 from 3 which is 2 Let's say you were subtracting 3-3=? to do this we take 3 away from 3 now 3 is 0 so our answer is 0 so there are 3 different problems we can make with 3 we could make more but I'm just telling the basics Hope I Helped Bye :)
Explanation:
The molar mass of the compound potassium nitrate, KNO3 is equal to 101.1032 g/mol. Then, we determine the number of moles present in the given amount,
n = 11.75g / (101.1032 g/mol) = 0.116 mol
Then, molarity is calculated by dividing the number of moles by the volume of the solution. The answer is therefore 0.058 M.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
It begins with heating of phenol and formaldehyde in the presence of a catalyst such as hydrochloric acid, zinc chloride, or the base ammonia. This creates a liquid condensation product, referred to as Bakelite A, which is soluble in alcohol, acetone, or additional phenol.
