Answer:
b=200
l=100
Explanation:
so since it is a square all sides would be equal
when you cut the square vertically the dimensions will change
the bredth would be same that is 200
but the length would be half that is 100
Answer:
<em />
<em>a) Balanced chemical equation:</em>
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<em> </em><em />
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<em>b) Theoretical yield:</em>
c) % yield:
Explanation:
The complete question is:
<em>In a particular reaction 6.80g of dinitrogen trioxide gas (N₂0₃) was actually produced by reacting 8.75g of oxygen gas (O₂) with excess nitrogen gas (N₂)</em>
<em>a) Write a balanced chemical equation for the reaction. Be sure to include physical states in the equation.</em>
<em>b) Calculate the theoretical yield (in grams) of dinitrogen trioxide: Use dimensional analysis</em>
<em>c) Calculate the % yield of the product</em>
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<h2>Solution</h2>
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<em>a) Write a balanced chemical equation for the reaction. Be sure to include physical states in the equation.</em>
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<em> </em><em />
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Check the balance:
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Atom Left-handside Right-hand side
N 2×2=4 2×2=4
O 3×2=6 2×3=6
- Mole ratio: it is the ratio of the coefficients of the balanced equation
<em>b) Calculate the theoretical yield (in grams) of dinitrogen trioxide: Use dimensional analysis</em>
<em />
<u>1. Convert 8.75 g of O₂(g) to number of moles</u>
- number of moles = mass in grams / molar mass
- molar mass of O₂ = 15.999g/mol
- number of moles = 8.75g / 15.999 g/mol = 0.5469 mol O₂
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<u>2. Use dimensional analysis to calculate the maximum number of moles of N₂O₃(g) that can be produced</u>
<u>3. Convert to mass in grams</u>
- mass = number of moles × molar mass
- molar mass of N₂O3 = 76.01g/mol
- mass = 0.3646mol × 76.01g/mol = 27.7g N₂O3
<em>c) Calculate the % yield of the product</em>
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Formula:
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- %yield = (actual yield/theoretical yield)×100
Substitute and compute:
- % yield = (6.80g/27.7g)×100 = 24.5%
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Then the liqid would freeze. Freezeing point is the tempture of a liqid that would make it liquid.
The half-life of Th-232 is 1.405 × 10¹⁰ years
Time elapsed = 2.8 x 10⁹ years
Equation of radioactive decay:
A = A₀ = (1/2)^ t/t₁/₂
where A₀ is the initial amount, A is the amount after time t, t₁/₂ is the half file
The fraction of thorium-232 that remains in the rock after 2.8 billions years is,
A/A₀ = (1/2) ^ (2.8 x 10⁹/ 1.405 × 10¹⁰) = 0.871
Therefore, the percentage of thorium-232 in the rock that was dated at 2.8 billions year = 87.1%