Answer:
31.3grams
Explanation:
The amount of a radioactive element that remains, N(t), can be calculated by using the formula;
N(t) = N(o) × 0.5^{t/t(1/2)}
Where;
N(t) = mass of substance remaining
N(o) = mass of original substance
t(1/2) = half-life of substance
t = time elapsed
From the provided information; N(t) = ?, N(o) = 250g, t(1/2) = 1599 years, t = 4797 years
N(t) = 250 × 0.5^{4797/1599}
N(t) = 250 × 0.5^(3)
N(t) = 250 × 0.125
N(t) = 31.25g
N(t) = 31.3grams
Answer:
endothermic
Explanation:
more energy is being released and heat is being required to do the process.
Answer:
1. Isotope with mass number 39.
2. 39.02g/mol
Explanation:
1. The most common isotope is the isotope in the higher proportion, that is:
Isotope with mass number 39.
2. The average atomic mass is the sum of the masses times their abundance. For the element X:
Average atomic mass:
38*0.0967 + 39*0.7868 + 40*0.1134 + 41*0.0031
= 39.02g/mol
Chemical formula of the glucose: C₆H₁₂O₆
We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u
atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol
2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g---------------------- x
x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles
we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)
3)We calculate the number of molecules:
Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles-------- x
x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.
4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²
number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .
number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²
Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
