Answer:
1 million years
Explanation:
The half life of a radioactive isotope refers to the time it takes for the radioactive isotope to decay to half of its original amount.
The half life of a radioactive isotope is independent of the amount of starting material. Hence, whether the amount of starting material is small or large, the half life of the substance remains the same.
Hence the half life of 2 green marbles and 500,000 green marbles is 1 million years.
Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
Answer:
Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.
Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.
Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.
Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.
Re-weigh the crucible and contents once cold.
Calculation:
Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)
Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment
Calculate the number of moles of anhydrous copper(II) sulfate formed
Calculate the number of moles of water driven off
Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed
Write down the formula for hydrated copper(II) sulfate.
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Explanation:
HI
So, the formula for water is H2O
When you have the same amount of the reactants , hydrogen will be the limiting reactant.
Limiting reactant is the thing that runs out first.
