Recycling reduces the demand for new wood is correct
Answer:
The answer to your question is 7.4 moles of Aluminum
Explanation:
Data
moles of Al = ?
moles of Al₂O₃ = 3.7
Balanced chemical reaction
4 Al + 3 O₂ ⇒ 2 Al₂O₃
To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.
4 moles of Aluminum ----------------- 2 moles of Al₂O₃
x ----------------- 3.7 moles of Al₂O₃
x = (3.7 x 4) / 2
x = 14.8 / 2
x = 7.4 moles of Aluminum
Answer is: elements in group 1 will lose electrons to obtain a noble gas structure. They will lose 1 electron.
For example ₃Li 1s²2s¹ will lose one electron from 2s oribtal to obtain helium structure ₂He 1s².
Or sodium ₁₁Na 1s²2s²2p⁶3s¹ will also lose one electron to obtain neon structure ₁₀Ne 1s²2s²2p⁶.
Answer:
molar mass M(s) = 65.326 g/mol
Explanation:
- M(s) + H2SO4(aq) → MSO4(aq) + H2(g)
∴ VH2(g) = 231 mL = 0.231 L
∴ P atm = 1.0079 bar
∴ PvH2O(25°C) = 0.03167 bar
Graham´s law:
⇒ PH2(g) = P atm - PvH2O(25°C)
⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm
∴ nH2(g) = PV/RT
⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))
⇒ nH2(g) = 9.1082 E-3 mol
⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))
⇒ n M(s) = 9.1082 E-3 mol
∴ molar mass M(s) [=] g/mol
⇒ molar mass M(s) = (0.595 g) / (9.1082 E-3 mol)
⇒ molar mass M(s) = 65.326 g/mol
Answer:
For example, the atomic radius of the metal zirconium, Zr, (a period-5 transition element) is 155 pm (empirical value) and that of hafnium, Hf, (the corresponding period-6 element) is 159 pm. ... The increase in mass and the unchanged radii lead to a steep increase in density from 6.51 to 13.35 g/cm3.
Explanation:
