PH + pOH = 14
pH + 0.253 = 14
pH = 14 - 0.253
pH = 13.747
[ H+] = 10 ^ -pH
[ H+ ] = 10 ^- 13.747
[ H+ ] = 1.790x10⁻¹⁴ M
hope this helps!
Answer:
91.26 g
Explanation:
Given data:
Mass of PF₃ = 180 g
Mass of F₂ required = ?
Solution:
Chemical equation:
P₄ + 6F₂ → 4PF₃
Moles of PF₃:
Number of moles = mass/ molar mass
Number of moles = 180 g/ 88 g/mol
Number of moles = 2.05 mol
Now we will compare the moles of PF₃ with F₂.
PF₃ : F₂
4 : 6
2.05 : 6/4×2.05 = 3.075
Mass of F₂:
Mass of F₂ = moles × molar mass
Mass of F₂ = 3.075 mol × 38 g/mol
Mass of F₂ = 116.85 g
If reaction yield is 78.1%:
116.85 /100 ×78.1 = 91.26 g
(i’m not very good at writing but use this as an idea)
changing the told would mess up the amount a daylight/nighttime because they’d be at a different angle from the sun
When carbon is burned in air carbon iv oxide gas is formed.
C (s) + O2 (g) = CO2(g) ΔH = - 393.5 kj/mol
The enthalpy change of the reaction is -393.5 j/mol which means that when one mole of carbon is completely burnt in air then 393.5 j of energy is evolved.
Thus, 1 mole = -393.5 j , then for 480 kj
= 480 × 1/393.5
= 1.2198 moles
1 mole of carbon iv oxide is equal to 44 g
thus, 1.2198 moles will be 1.2198 × 44 = 53.6712 g of CO2
Answer: Unit conversions give the same physical value, but are represented with different units.
Explanation:
