Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
<span>Antoine Lavoisier is the answer </span>
<span>We can use the heat
equation,
Q = mcΔT </span>
<span>
Where Q is the amount of energy transferred (J), m is
the mass of the substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature
difference (°C).</span>
According to the given data,
Q = 300 J
m = 267 g
<span>
c = ?
ΔT = 12 °C</span>
By applying the
formula,
<span>300 J = 267 g x c x
12 °C
c = 0.0936 J g</span>⁻¹ °C⁻<span>¹
Hence, specific heat of the given substance is </span>0.0936 J g⁻¹ °C⁻¹.
Answer:
P(H₂) = 0.8533 atm
Explanation:
n(CO) = 0.220 mole
n(H₂) = 0.350 mole
n(He) = 0.640 mole
_______________
∑ n = 1.210 moles
mole fraction => X(H₂) = 0.350/1.210 = 0.2892
Dalton's Law of Partial Pressures => P(H₂) = X(H₂)·P(ttl) = 0.2892(2.95 atm) = 0.8533 atm
Answer:
in a liquid, particles will flow or glide over one another, but stay toward the bottom of the container.
Explanation:
the attractive forces between particles are strong enough to hold a specific volume but not strong enough to keep the molecules sliding over each other.
