hydrogen and helium and the most abundant elements in a nebula.
Answer:

Explanation:
Question 7.
We can use the Combined Gas Laws to solve this question.
a) Data
p₁ = 1.88 atm; p₂ = 2.50 atm
V₁ = 285 mL; V₂ = 435 mL
T₁ = 355 K; T₂ = ?
b) Calculation

Question 8. I
We can use the Ideal Gas Law to solve this question.
pV = nRT
n = m/M
pV = (m/M)RT = mRT/M
a) Data:
p = 4.58 atm
V = 13.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 385 K
M = 46.01 g/mol
(b) Calculation

Answer: they all have 7 electrons they have different atom sizes
Explanation: hope it helps
The correct answer is that last one
Unit ppm stands for parts per million. in terms of mass, ppm is equivalent to mg/kg.
since 1 kg is 10⁻⁶ mg, 1 kg is equivalent to million mg.
therefore mg/kg is also ppm.
there are 1.3 g of Ca ions in 3100 kg
if 3100 kg contains - 1.3 g of Ca
then 1 kg contains - 1.3 g / 3100 kg
then Ca ions - 0.42 x 10⁻³ g/kg
Ca ion concentration - 0.42 mg/kg
therefore Ca ion concentration is 0.42 ppm