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White raven [17]
3 years ago
10

What is the mass of 4.76 moles of Na3Po4?

Chemistry
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

780 g

Explanation:

Step 1: Given data

Moles of sodium phosphate: 4.76 moles

Step 2: Calculate the mass corresponding to 4.76 moles of sodium phosphate

To convert moles to mass, we need a conversion factor. In this case, we will use the molar mass of sodium phosphate, which is 163.94 g/mol.

4.76 mol × 163.94 g/mol = 780 g

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I have a question for us gals, is it hard for all of you when you try to get your crushes attention and he/she just keeps ignori
Volgvan

Answer:

Yes it is hard. For some of us. I already gave up, but now hes coming back. Id say sometimes its useless to get his attention if he never even pays attention to you anyways. It sucks because its so hard to get then to notice you for once.

Explanation:

4 0
3 years ago
Some animals that live in extremely cold environments protect themselves from freezing by increasing the molality of certain che
lara [203]
  • Molality is defined as the number of moles of solute dissolved per 1000g(1kg) of solvent.
  • It is represented by m

\\ \sf \longmapsto m=\dfrac{Moles\:of\:solute}{Wt\:of\:solvent(in\:g)}\times 1000

8 0
3 years ago
Se ha añadido un evaporador para una alimentación de 11500 kg/dia de zumo de pomelo de forma que evapore 3000 kg/dia de agua por
mamaluj [8]

Answer:

37 %.

Explanation:

¡Hola!

En este caso, para el problema descrito, conocemos la corriente de entrada y la de salida del agua, por lo que podemos obtener el flujo de la corriente que contiene el zumo a la salida una vez el agua fue evaporada:

F_{sol}=11500kg/dia-3000kg/dia=8500 kg/dia

Luego, por medio de un balance de zumo de limón en el evaporador en el cual la cantidad que entra es igual a la que sale con sus respectivas concentraciones:

x_z^{entra}*11500kg/dia=x_z^{sale}*8500kg/dia

Como la concentración del zumo a la salida es del 50 % (0.50), la de entrada es:

x_z^{entra}=\frac{x_z^{sale}*8500kg/dia}{11500kg/dia} =\frac{0.50*8500kg/dia}{11500 kg/dia}\\ \\x_z^{entra}=0.37

Que es igual al 37%.

¡Saludos!

6 0
4 years ago
Why calcium oxide and magnesium oxide used as soil treatment ​
BartSMP [9]

Answer:

Chemically speaking, lime refers only to calcium oxide

(CaO); however, in common usage the term includes the calcination products of calcitic and dolomitic limestones. Calcitic (high-calcium) limes are produced by calcination of

calcareous materials (e.g., calcitic limestone, calcite,

oyster shells, and chalk) containing from 95 to 99 percent

calcium carbonate (CaCO^). Dolomitic limes are produced from

dolomitic limestone or dolomite which contains from 30 to ^+0

percent magnesium carbonate (MgCO^), the rest being calcium

carbonate.

At atmospheric pressure, calcite in limestone decomposes

at approximately 900°C to form CaO and COg. The decomposition of dolomite, CaMg is a two-stage process. At

temperatures between 650°C to 750°C dolomite decomposes to

form MgO, CO^ and CaCO^. It is necessary to raise the temperature to 900°C to decompose the CaCO^ (15, 35)» This

phenomenon is extremely important, as is shown later.

Various investigators have studied the effects of stone

size, temperature, and time of calcination of commercial

Explanation:

4 0
3 years ago
Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -
kari74 [83]

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)  \Delta H=-2220.1kJ/mol

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]

\Delta H_{C_3H_8}=72.4kJ/mol

Therefore, the heat of formation of propane is 72.4 kJ/mol

7 0
3 years ago
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