1) Calcium carbonate comprises 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (which is 40.0%)!
2) The mass fraction mentioned is superfluous information.
3) The resulting solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
Answer:
Explanation:
Assuming all calculations occur at standard pressure and a temperature of -1.72°C :
Where
is the number of moles of hydrogen
is the mass of hydrogen
is the density of hydrogen
Answer:
B) Hyperbolic curve; substrate saturation
Explanation:
Enzymatic kinetics examines the rates of reactions catalyzed by enzymes. These studies offer insights into the mechanism of the catalytic reaction and enzyme specificity. Determining the reaction rate facilitated by an enzyme is generally straightforward, as purification or isolation of the enzyme is frequently unnecessary. Measurements are taken under optimal conditions for pH, temperature, and the presence of cofactors, utilizing saturating substrate concentrations. Under these circumstances, the observed reaction rate is the maximum velocity (Vmax). The rate can be measured by monitoring either product formation or substrate consumption.
Following the rate of product formation (or substrate consumption) over time yields the so-called reaction progress curve, or merely, reaction kinetics. This reacts as a hyperbolic curve
Answer:
The glycerol solution has a molality of 2.960×10^-2 mol/kg.
Explanation:
Calculating the moles of glycerol involves the formula: Moles = Molarity × Volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles.
To find the mass of water, use: Mass = Density × Volume = 0.9982 g/mL × 998.7 mL = 996.90 g, which converts to 0.9969 kg.
The formula for molality is: Molality = Moles of solute/Mass of solvent (in kg) = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg.
The formula for molality is
Given: A solution's molality equals 2.25 m
The weight of the solvent is 30 g, which is the same as 0.030 kg
Molecular weight of AlCl3 is 133.34 g/mol
Thus, we have the equation, 2.25 =
Therefore, the weight of the solute (g) becomes 9.00 g
Hence, <span>
9.00 g of AlCl3 is necessary for creating a 2.25m solution in 30.0 g of water</span>
