Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
Answer:
2.78 x 10²³
Explanation:
1 mole contains 6.02 x 10²³ hydrogen atoms => 0.46 mole contains 0.46(6.02 x 10²³) hydrogen atoms or 2.78 x 10²³ atoms.
Caution => When to use H vs H₂ => This problem is specific for 'hydrogen atoms' but some may simply say hydrogen. In such cases use H₂ or 'molecular hydrogen' is the focus. it's a matter of semantics, H vs H₂.
☃️ Chemical formulae ➝ 
<h3>
<u>How to find?</u></h3>
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
<h3>
<u>Solution:</u></h3>
Atomic weight of elements:
Ca = 40
C = 12
O = 16
❍ Molecular weight of
= 40 + 12 + 3 × 16
= 52 + 48
= 100 g/mol
❍ Given weight: 10 g
Then, no. of moles,
⇛ No. of moles = 10 g / 100 g mol‐¹
⇛ No. of moles = 0.1 moles
☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>
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solubility will increase, since it is basic in nature, on acidifying solubility increases.
being basic on acidification, the solubility increases.
is alkaline in nature, it gets neutralized and there is no increase in acidity.
is a salt and it shows no effect on acidification.
is a salt but insoluble in water and shows no effect on acidification.
is strongly basic in nature, on acidifying the solubility increases.