Answer:
The answer to your question is V = 19.9 L
Explanation:
Data
mass of Al = 8.60 g
volume of H₂ = ?
Balanced chemical reaction
2Al + 6HCl ⇒ 2AlCl₃ + 3H₂
1.- Use proportions to calculate the mass of H₂ produced
Atomic mass of Al = 2 x 27 = 54 g
Atomic mass of H₂ = 6 x 1 = 6 g
54 g of Al -------------------- 6 g of H₂
8 g of Al ------------------- x
x = (8 x 6) / 54
x = 48 / 54
x = 0.89 g of H₂
2.- Calculate the volume of H₂
- Convert the H₂ to moles
1 g of H₂ --------------- 1 mol
0.89 g ---------------- x
x = 0.89 moles
1 mol of H₂ -------------- 22.4 l
0.89 moles ------------- x
x = (0.89 x 22.4) / 1
x = 19.9 L
Answer:
Explanation:
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)
ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%20%5B1%28-1675.7%29%20%2B%202%280%29%5D%20-%20%5B2%280%29%20-%201%28-824.3%29%5D%5C%5C%26%20%3D%20%26%20-1675.7%20%2B%20824.3%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20change%20is%20%7D%20%5Clarge%20%5Cboxed%7B%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%7D)
Answer:
Measuring a pencil in meters would be very difficult, as a single meter is much longer than one pencil. Also, measuring a hallway in millimeters would be very difficult considering how small millimeters are in comparison to a hallway. However, if you switch these two then they would work very well.
Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
POH = - log [ OH⁻ ]
pOH = - log [ 1 x 10⁻⁹ ]
pOH = 9
Answer C
hope this helps!
