Answer:
51 J
Explanation:
The air inside a bicycle tire pump has 27 joules of heat conducted away. By convention, when heat is released, it takes the negative sign, so Q = -27 J.
77.9 joules of work done are being done on the air inside a bicycle tire pump. By convention, when work is being done on the system, it takes the positive sign, so W = 77.9 J
We can calculate the change in the internal energy (ΔU) using the following expression.
ΔU = Q + W
ΔU = (-27 J) + 77.9 J
ΔU = 51 J
Answer is: not enough <span>colorless syrupy liquid.
</span>n(H₂SO₄) = 1,2 mol.
M(H₂SO₄) = 2Ar(H) + Ar(S) + 4Ar(O) · g/mol.
M(H₂SO₄) = 2·1 + 32 + 4·16 · g/mol.
M(H₂SO₄) = 98 g/mol.
m(H₂SO₄) = n(H₂SO₄) · M(H₂SO₄).
m(H₂SO₄) = 1,2 mol · 98 g/mol.
m(H₂SO₄) = 117,6 g needed.
100 g is less that 117,6 g.
Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.
He worked with large numbers of plants.
