A reaction was conducted between barium nitrate and sodium phosphate. 3 Ba(NO3)2 + 2 Na3PO4 → 6 NaNO3 + Ba3(PO4)2 What is the pe

Although I may not be the smartest, I can definitely answer.

This represents a chemical change because the substances' chemical identities were altered. The fizzing was a clear sign, and the temperature increase was another indicator of the reaction.

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:

The ratios arranged in ascending order are; The ratio of the mass of Y to X in XY2 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY3 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY4 divided by the mass of Y to X in XY

1) Mass ratio = 3

2) Mass ratio = 2

3) Mass ratio = 4

Explanation:

Comprehensive calculations are displayed in the attachment.

Answer: Pentane C5H12

Explanation:

The boiling point is defined as the temperature at which a liquid's vapor pressure matches the external pressure, causing the liquid to turn into vapor.

This compound is likely Pentane, represented as C5H12, since its boiling point falls between that of Butane, with the formula C4H10, and Hexane, with the formula C6H14.

A triprotic acid is a type of Arrhenius acid that has the ability to donate three protons per molecule during dissociation in aqueous solutions. Thus, the chemical reaction, as outlined in the question, at the third equivalence point, can be expressed as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R denotes the counter ion of the triprotic acid. Consequently, the ratio of reacted acid to base at this point is 1:3.
The moles of NaOH are calculated as 0.106M*0.0352L = 0.003731 mole. Therefore, the amount of H3R is 0.003731mole/3=0.001244mole.
Subsequently, the molar mass of the acid can be determined: 0.307g/0.001244mole=247 g/mol.