Answer:
0.508 mole
Explanation:
NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.
The number of mole present in 78.2 g of CCl₄ can be obtained as follow:
Mass of CCl₄ = 78.2 g
Molar mass of CCl₄ = 12 + (35.5×4)
= 12 + 142
= 154 g/mol
Mole of CCl₄ =?
Mole = mass / molar mass
Mole of CCl₄ = 78.2 / 154
Mole of CCl₄ = 0.508 mole
Therefore, 0.508 mole is present in 78.2 g of CCl₄
Answer:If we have [H+][OH-]= Kw = 1.0 x 10^-14
Then [H+]= Kw/ [OH-]= 1.0x 10^-14/ 1 x 10^-11 =1 x 10^-3 mol/L
And here is the solution - as you can see it is an acidic one :
pH = - log [H+]= - log 1 x 10^-3 = 3 < 7
Explanation:
Answer:
Protein Concentration is 2.82mg/L
Explanation:
According to Beer-Lambert's Law, Absorbance is directly proportional to the concentration.
However, the concentration of a solution can be determined from a calibration curve, in which Absorbance is plotted on the y-axis and the Concentration on the x-axis.
Plotting the best line, the equation of line is used
y = mx + c
where y is absorbance = 0.150
m is slope = 0.0163
x is concentration
c is intercept = 0.104
inserting the values from the question
y = mx + c
0.150 = 0.0163x + 0.104
0.0163x = 0.150 - 0.104
0.0163x = 0.046
Divide both sides by 0.0163
0.0163x/0.0163 = 0.046/0.0163
x = 2.82
Concentration of protein = 2.82 mg/L
Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s
We will get the molality from this formula:
Molality = no.of moles of solute / Kg of solvent
So first we need the no.of moles of KNO3 = the mass of KNO3 / molar mass of KNO3
no.of moles of KNO3 = 175 / 101.01 = 1.73 mol
By substitution in the molality formula:
∴ molality = 1.73 / (750/1000) = 2.3 Molal
