Answer:
0.053moles
Explanation:
Hello,
To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,
V = kN, k = V / N
V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx
V1 = 1.7L
N1 = 0.070mol
V2 = 1.3L
N2 = ?
From the above equation,
V1 / N1 = V2 / N2
Make N2 the subject of formula
N2 = (N1 × V2) / V1
N2 = (0.07 × 1.3) / 1.7
N2 = 0.053mol
The number of moles of gas in his lungs when he exhale is 0.053 moles
The pH of the solution after adding 0.150 moles of solid LiF is 3.84
<u>Explanation:</u>
We have the chemical equation,
HF (aq)+NaOH(aq)->NaF(aq)+H2O
To find how many moles have been used in this
c= n/V=> n= c.V
nHF=0.250 M⋅1.5 L=0.375 moles HF
Simillarly
nF=0.250 M⋅1.5 L=0.375 moles F
nHF=0.375 moles - 0.250 moles=0.125 moles
nF=0.375 moles+0.250 moles=0.625 moles
[HF]=0.125 moles/1.5 L=0.0834 M
[F−]=0.625 moles/1.5 L=0.4167 M
To determine the problem using the Henderson - Hasselbalch equation
pH=pKa+log ([conjugate base/[weak acid])
Find the value of Ka
pKa=−log(Ka)
pH=−log(Ka) +log([F−]/[HF]
pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)
pH=-log(3.5 x 10 ^4)+log(4.996)
pH= -4.54+0.698
pH=-(-3.84)
pH=3.84
The pH of the solution after adding 0.150 moles of solid LiF is 3.84
I'm sure that to calculate the freezing point depression <span>subtract</span> solution's freezing point and the freezing point of it's pure solvent. According to the formula.
Answer:
cubic measurements
Explanation:
i.e. cubic meter or cubic centimeter
Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.