Answer:
6.56 × 10²² atoms Xe
Explanation:
Step 1: Find conversions
Molar Mass Xe - 131.29 g/mol
Avagadro's Number: 6.022 × 10²³
Step 2: Use Dimensional Analysis
= 6.55911 × 10²² atoms Xe
Step 3: Simplify
We have 3 sig figs.
6.55911 × 10²² atoms Xe ≈ 6.56 × 10²² atoms Xe
Answer:
The given element is Radon because its atomic weight is 222 amu.
Explanation:
Given data:
Percentage of A-219 = 13.92%
Percentage of B-222 = 72.16%
Percentage of C-225 = 13.92%
Atomic weight of element = ?
Solution:
Average atomic mass = (abundance of A isotope × its atomic mass) +(abundance of B isotope × its atomic mass) + (abundance of C isotope × its atomic mass) / 100
Average atomic mass = (13.92×219)+(72.16×222) + (13.92×225)/100
Average atomic mass = 3048.48 + 16019.52 +3132/ 100
Average atomic mass = 22200 / 100
Average atomic mass = 222 amu.
The given element is Radon because its atomic weight is 222 amu.
Answer: 2NOBr(g) ⇌ 2NO(g) + Br2(g)
Explanation: For volume changes in equillibrium, the following are to be taken into consideration:
- Volume changes have no effect on equillibrium system that contains solid or aqueous solutions.
- An increase in volume of an equilibrium system will shift to favor the direction that produces more moles of gas.
- A decrease in volume of an equilibrium system will shift to favor the direction that produces less moles of gas.
- Volume changes will have no effect on the equillibrium system if there is an equal number of moles on both sides of the reaction.
2NOBr(g) ⇌ 2NO(g) + Br2(g) is the equillibrium system because there are more moles of products,therefore an increase in the volume of the reaction will shift to the right and produce more moles of products. Also both reactants and products exist in the gaseous state and does not have equal number of moles.
A: The atoms in the sand were moving more quickly
First find the molar mass (mm) for Methane--> CH4 = 12+(1×4)
= 16 grams/mole CH4 --> now flip this:
So the = # of moles (n)
= 12.5g CH4 × 1 mole CH4 / 16g CH4
n = 0.78 moles
P = 1.30 atm, T = 40C = 273K + 40C
= 313K, V = ??
PV = nRT --> V = nRT/P
V = (0.78)(.082)(313) / 1.30
V = 20.1 / 1.30 = 15.4 L
