Answer: exactly 82 protons
Explanation:
Since lead has 82 protons the number of neutrons in an atom of lead-204 is 204-82=122. Similarly, lead-206 has 124 neutrons, lead-207 has 125 neutrons and lead-208 has 126 neutrons.
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Find the mass of the empirical formula.
You must be given a sample of some kind to calculate the weight or know how many moles are present. Then you figure out what one mol would be. The key step is multiplying the empirical formula numbers by what it takes to make 1 mol.
It would be clearer if we were working from some choices.
Greetings!
To find the empirical formula you need the relative atomic mass of each element!
Li = 6.9
C = 12
O = 16
You can simply change the percentages into full grams
Li = 18.8g
C = 16.3g
O = 64.9
Then you use this to find the Number of moles = amount in grams / atomic mass
Li = 18.8 ÷ 6.9 = 2.7246
C = 16.3 ÷ 12 = 1.3583
O = 64.9 ÷ 16 = 4.0562
Then divide each number of moles by the smallest value:
Li = 2.7246 ÷ 1.3583 = 2.0
C = 1.3583 ÷ 1.3583 = 1
O = 4.0562 ÷ 1.3583 = 2.9 ≈ 3
So that means that there are 2 Li, 1 C, and 3 O
Empirical formula would be:
Li₂CO₃
Hope this helps!
Answer:
11.6 mol O₂
Explanation:
- C₇H₁₆ + 11 O₂ → 7 CO₂ + 8 H₂O
In order to solve this problem we need to <u>convert moles of carbon dioxide (CO₂) into moles of oxygen gas (O₂)</u>. To do so we'll use a conversion factor containing the <em>stoichiometric coefficients</em> of the balanced reaction:
- 7.4 mol CO₂ *
= 11.6 mol O₂
Answer:
t = 7.58 * 10¹⁹ seconds
Explanation:
First order rate constant is given as,
k = (2.303
/t) log [A₀]
/[Aₙ]
where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>
[A₀] = 615 calories;
[Aₙ] = 615 - 480 = 135 calories
k = 2.00 * 10⁻²⁰ sec⁻¹
substituting the values in the equation of the rate constant;
2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)
(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)
t = 2.303 / 3.037 * 10⁻²⁰
t = 7.58 * 10¹⁹ seconds