Mass Molar of

Ca = 3*40 = 120 amu
P = 2*31= 62 amu
O = (16*4)*2 = 64*2 = 128 amu
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Mass Molar of

= 120 + 62 + 128 = 310 g/mol
Therefore: <span>What is the gram formula mass of Ca3(PO4)2 ?
</span>Answer:
310 grams
1 mole Hg --------------- 6.02x10²³ atoms
?? moles Hg ------------ 1.30x10⁷ atoms
(1.30x10⁷) x 1 / 6.02x10²³ => 2.159x10⁻¹⁷ moles
Answer:

Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure

(c) Convert the pressure to atmospheres

Just breaking stuff so yea that’s it