Response:
of 
can be found in 39.5 grams of .
Clarification:
Atomic weights: P= 31, F= 19,
The molar mass equals 1 atomic weight of P + 5 atomic weights of
F= 31+5 × 19
= 31+95
=126 g/mole
The number of moles in 39.5 gm of
equals 
= 
=0.3134 moles
1 mole of any substance encompasses
0.3131 moles comprises 0.3134
Thus, of can be found in 39.5 grams of .
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
Respuesta:
0.16 M
Explicación:
Teniendo en cuenta:
O sea,
Dado que:
Para 
:
Molaridad = 0.200 M
Volumen = 20.0 mL
Convierte mL a L:
1 mL = 10⁻³ L
Entonces, volumen = 20.0×10⁻³ L
Los moles de son:
Moles de = 0.004 moles
Para 
:
Molaridad = 0.400 M
Volumen = 30.0 mL
Convertimos mL a L:
1 mL = 10⁻³ L
Volumen = 30.0×10⁻³ L
Entonces, los moles de son:
Moles de = 0.012 moles
Según la reacción:
1 mol de reacciona con 1 mol de
Por lo tanto,
0.012 mol de reacciona con 0.012 mol de
Moles disponibles de = 0.004 mol
El reactivo limitante es el que está en menor cantidad, entonces es el limitante (0.004 < 0.012).
La formación del producto depende del reactivo limitante, así que,
1 mol de reacciona con 1 mol de y produce 1 mol de 
0.004 mol de reacciona con 0.004 mol de y genera 0.004 mol de
Los moles restantes de son: 0.012 - 0.004 = 0.008 mol
El volumen total es 20 + 30 mL = 50 mL = 0.050 L
Por lo que la concentración del ion bario, 
, después de la reacción es:
To solve for density, you can use the formula--> Density= PM/ RT, where P stands for pressure, M for molar mass, R represents the gas constant, and T is temperature.
P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k
Thus, the density calculation becomes: density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L
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.
