0.25 moles of CO2 is present in 11 grams of CO2.
Explanation:
A mole represents the number of chemical entities in an element or molecule.
Number of moles of an element or molecule is determined by the formula:
The Number of moles (n) = weight of the atom given ÷ atomic or molecular weight of the one mole of the element or molecule.
Themolar mass of one mole of carbon dioxide is:
12+ ( 16×2)
= 44 gram/mole
The given weight is 44 grams of carbon dioxide.
Putting the values in the equation,
n= 11 gms÷44 gms/ mole
n = 0.25 mole
The F atom . Oxygen has a -2 charge and fluorine has a +1 charge
Answer:
C₁₁H₁₂NO₄
Explanation:
In order to determine the empirical formula of doxycycline, we need to follow a series of steps.
Step 1: Determine the centesimal composition
C: 59.5 mg/100 mg × 100% = 59.5%
H: 5.40 mg/100 mg × 100% = 5.40%
N: 6.30 mg/100 mg × 100% = 6.30%
O: 28.8 mg/100 mg × 100% = 28.8%
Step 2: Divide each percentage by the atomic mass of the element
C: 59.5 /12.0 = 4.96
H: 5.40/1.00 = 5.40
N: 6.30/14.0 = 0.450
O: 28.8/16.0 = 1.80
Step 3: Divide all the numbers by the smallest one
C: 4.96/0.450 = 11
H: 5.40/0.450 = 12
N: 0.450/0.450 = 1
O: 1.80/0.450 = 4
The empirical formula of doxycycline is C₁₁H₁₂NO₄
Answer:
10. mol NaOH
Explanation:
2.5 M = 2.5 mol/L
2.5 mol/L * 4 L = 10. mol
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
