The mass of 254 mL of water is 254 g. Since the density of water is 1g/mL, we can simply multiply the density 1g/mL by 254 mL of water and get 254 g as our answer. Since mL is in the numerator and denominator, mL cancels out and we are left with g only.
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Write an balance the equation
Na2O + H2O -> 2 NaOH
Calculate the molecular mass of Na2O and NaOH from the atomic mass from the periodic table.
Na = 23
O=16
H=1
Na2O = 23 * 2 + 16 = 62
NaOH = 23+16+1= 40
For the stoichiometry of the reaction one mole of Na2O = 62g produce two mol of NaOH = 2* 40= 80 g
120 g Na2O x 80g NaOH / 62g Na2O=
154.8 g NaOH
Answer:
The answer is b. The number of collisions of gas particles increases
