The empirical formula :
C₁₀H₁₆N₄SO₇
<h3>Further explanation</h3>
Given
6.4 g sample
Required
The empirical formula
Solution
mass C :
= 12/44 x 8.37 g
= 2.28
mass H :
= 2/18 x 2.75 g
= 0.305
mass N = 1.06
mass S :
= 32/64 x 1.23
= 0.615
mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g
Mol ratio :
= C : H : N : S : O
= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16
= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019
= 10 : 16 : 4 : 1 : 7
The empirical formula :
C₁₀H₁₆N₄SO₇
Molarity=Moles of solute/Volume of solution in L
So
- 0.56M=moles/2.5L
- moles=0.56(2.5)
- moles of Iodine=1.4mol
Mads of Iodine
- Moles(Molar mass)
- 1.4(126.9)
- 177.66g
Answer:
Option 4 ) 1-butyne
Explanation:
In organic chemistry, you should use IUPAC convention in order to name an organic compound. First of all, you should identify the lenght of the organic chain, for this case, you have 5 C atoms, but in fact, you have a triple bond (which would be a substitute: ethynil-) as a substitute, so the main skeleton would have 4 C atoms (a butane)
Then, you start by numbering carbon N° 1 as the one that has the substitute (triple bound)-starting from the right, it would be the second C):
CH₃-CH₂-CH₂-C≡CH
Which will finally leads us to 1-butyne
Answer:
Keq = [CO₂]/[O₂]
Explanation:
Step 1: Write the balanced equation for the reaction at equilibrium
C(s) + O₂(g) ⇄ CO₂(g)
Step 2: Write the expression for the equilibrium constant (Keq)
The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:
Keq = [CO₂]/[O₂]
