Answer:
Option D. 0.115 M
Explanation:
The following data were obtained from the question:
Mass of CuSO4 = 36.8 g
Volume of solution = 2 L
Molar mass of CuSO4 = 159.62 g/mol
Molarity of CuSO4 =..?
Next, we shall determine the number of mole in 36.8 g of CuSO4.
This can be obtained as shown below:
Mass of CuSO4 = 36.8 g
Molar mass of CuSO4 = 159.62 g/mol
Mole of CuSO4 =.?
Mole = mass /Molar mass
Mole of CuSO4 = 36.8 / 159.62
Mole of CuSO4 = 0.23 mole
Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:
Mole of CuSO4 = 0.23 mole
Volume of solution = 2 L
Molarity of CuSO4 =..?
Molarity = mole /Volume
Molarity of CuSO4 = 0.23 / 2
Molarity of CuSO4 = 0.115 M
Therefore, the molarity of the CuSO4 solution is 0.115 M.
<span>The equation that describes the problem is Fe(NO3)3(aq) + 3NaOH(aq) ---> Fe(OH)3(s) + 3 NaNO3(aq)
The Net ionic equation is written as follows:
Fe^3(aq) + 3NO3-(aq) + 3Na+(aq) + 3OH-(aq) ---> Fe(OH)3(s) + 3Na+(aq) + 3NO^3-(aq)</span>
Answer:
Explanation:
El hielo encierra la primavera antes de espolvorear la sal.
Cuando se rocía sal sobre el hielo, disminuye el punto de fusión del hielo 32 ° F a un poco por debajo de 32 °, por lo tanto, se acumula.
A medida que el hielo se vuelve a congelar, encierra la primavera
