Answer:
4.90 x 10 24 atoms
Explanation:
the 24 is the exponent for the 10
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
Answer:
In the 1H NMR spectrum of ethanol three different signals are observed, this is due to the existence of 3 types of hydrogens with different chemical environment. Hydrogens A (3.57 ppm) are more screened than C (1.10 ppm) due to the presence of oxygen (electonegative atom that removes electron density). The chemical environment of hydrogen B (4.78 ppm), attached directly to oxygen, is also different by resonating at a frequency different from the previous ones.
The hydroxylic hydrogen produces a singlet, the pair of carbon hydrogens one give rise to a quadruplet and the three hydrogens of carbon two produce a triplet.
Explanation:
Henry's law constant for oxygen is 0,0013 mol/L·<span>atm. Air has 21,0% oxygen.
concentration of oxygen at 1 atm: 0,0013 mol/L</span>·atm · 0,21 · 1 atm = 0,000273 mol/l.
concentration of oxygen at 1 atm: 0,0013 mol/L·atm · 0,21 · 0,892 atm = 0,000243 mol/l.
difference in concentration: 0,000273 - 0,000243 = 0,00003 mol/L.
n(oxygen) = 0,00003 mol/L · 4,40 L = 0,000132 mol.
Answer:
a) K = [ CO2(g) ]
⇒ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
Explanation:
a) CaCO3(s) ↔ CaO(s) + CO2(g)
⇒ K = [ CO2(g) ]
∴ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) H2(g) + F2(g) ↔ 2 HF(g)
⇒ K = [ HF(g) ] ² / [ F2(g) ] * [ H2(g) ]
⇒ Kp = PHF² / PF2 * PH2
for ideal gas:
PV = RTn
⇒ P = n/V RT = [ ] RT
⇒ Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same.
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
