Enter the appropriate symbol for an isotope of phosphorus-32 corresponding to the isotope notation AZX. Express your answer as a

The accurate answer is 33.8 g.

Ca3(PO4)2 is the correct formula.

Noble gas notation serves as a condensed form of indicating electron configurations. This notation employs the symbol for the preceding noble gas in the electron configuration of an element. For antimony, the noble gas prior is Kr, which means Xe is not used in its electron configuration. Similarly, for radium, the prior noble gas is Rn, whereas, for uranium, it is also Rn. However, for cesium, the preceding noble gas is Xe, thus it is utilized in the noble gas notation for Sb, specifically written as: Cs: [Xe] 6s.

Answer: cesium

Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

<span>128 g/mol Applying Graham's law of effusion, we can utilize the formula: r1/r2 = sqrt(m2/m1) where r1 = effusion rate of gas 1 r2 = effusion rate of gas 2 m1 = molar mass of gas 1 m2 = molar mass of gas 2 Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998. We can now insert the known values into Graham's equation to find m2. r1/r2 = sqrt(m2/m1) 2/1 = sqrt(m2/31.998) 4/1 = m2/31.998 Thus, we find m2 to be 127.992. Rounding to three significant figures yields 128 g/mol</span>