Electrons are orbiting the nucleus in the fxed way paths located in solid sphere
<span>Answer:
(16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m
a.)
(3.9704 m) x (1.86 °C/m) = 7.38 °C change
0.00°C - 7.38 °C = - 7.38 °C
b.)
(3.9704 m) x (0.512 °C/m) = 2.03 °C change
100.00°C + 2.03 °C = 102.03 °C</span>
Answer:
Average atomic mass of carbon = 12.01 amu.
Explanation:
Given data:
Abundance of C¹² = 98.89%
Abundance of C¹³ = 1.11%
Atomic mass of C¹² = 12.000 amu
Atomic mass of C¹³ = 13.003 amu
Average atomic mass = ?
Solution:
Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass of carbon = (12.000×98.89)+(13.003×1.11) /100
Average atomic mass of carbon= 1186.68 + 14.43333 / 100
Average atomic mass of carbon = 1201.11333 / 100
Average atomic mass of carbon = 12.01 amu.
Answer:
119.7 mL.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas.
V is the volume of the container.
n is the no. of moles of the gas.
R is the general gas constant.
T is the temperature of the gas (K).
- For the same no. of moles of the gas at two different (P, V, and T):
<em>P₁V₁/T₁ = P₂V₂/T₂.</em>
- P₁ = 100.0 mmHg, V₁ = 1000.0 mL, T₁ = 23°C + 273 = 296 K.
- P₂ = 1.0 atm = 760.0 mmHg (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).
<em>∴ V₂ = (P₁V₁T₂)/(T₁P₂) </em>= (100.0 mmHg)(1000.0 mL)(273.0 K)/(296 K)(760.0 mmHg) = 121.4 <em>mL.</em>
