Answer:
a) 
b) 1657 €
Explanation:
Hola,
a) En esta cuestión analizaremos el millón de litros de agua anualmente, dado que este dato nos permite calcular el calor necesario para calentar dicha cantidad, considerando que la densidad del agua es de 1 kg/L:
A continuación, utilizamos la entalpía de combustión del metano para determinar la cantidad en kilogramos necesaria, ya que la energía calórica perdida por el metano es equivalente a la energía obtenida por el agua:
b) En este supuesto, tenemos que, bajo condiciones normales de 1 bar y 273 K, el precio de 1 metro cúbico de metano es 0,45 €, lo que nos permite calcular las moles de metano en esas condiciones:
En consecuencia, los kilogramos de metano que se obtienen por 0,45 € son:
Finalmente, usando regla de tres:
0.715 kg ⇒ 0.45 €
2630 kg ⇒ X
X = (2630 kg x 0.45 €) / 0.715 kg
X = 1657 €
Regards.
Answer:
The molality is 1.15 m.
Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.
Calculate moles of H₂SO₄ from molarity:
C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles
Mass of solvent (water) based on density:
m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg
Therefore, molality is:
m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m
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
In a water molecule, the sharing of electrons occurs between the oxygen and hydrogen atoms within covalent bonds; however, this sharing is unequal. The oxygen atom holds a stronger pull on the electrons compared to the hydrogen atoms in the bond.
Answer:
Please review the following responses
Explanation:
1) A solution of 100. mL contains 19.5 g of NaCl (3.3M)
2) 100. mL of NaCl solution at 3.00 M (3 M)
3) A solution of 150. mL holds 19.5 g of NaCl (2.2 M)
4) The concentrations of beakers 1 and 5 are identical (1.5M)
Molar mass of NaCl = 23 + 36 = 59 g
For beaker number 3:
59 g -------------- 1 mol
19.5 g ------------- x
x = 19.5 x 1/59 = 0.33 mol
Molarity (M) = 0.33 mol/0.150 l = 2.2 M
For beaker number 4:
Molarity (M) = 0.33mol/0.10 l = 3.3 M
For beaker number 5:
Molarity (M) = 0.450/0.3 = 1.5 M
