Simple dimensional analysis.
okay so youll need a periodic table to look up the molar mass. youll be given either an amount of grams or moles.
Mixture......................
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
The answer is: 27 grams of aluminium.
Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.
n(H₂) = 1.5 mol; amount of hydrogen.
Form chemical reaction: n(Al) : n(H₂) = 2 : 3.
n(Al) = 2 · 1.5 mol ÷ 3.
n(Al) = 1.0 mol; amount of aluminium.
m(Al) = n(Al) · M(Al).
m(Al) = 1 mol · 27 g/mol.
m(Al) = 27 g; mass of aluminium.