Suppose you held a lighted match to a solid hunk of wood and another match to a pile of wood shavings. Which form of wood will c

Answer:

The correct answer is "Speed".

Explanation:

• An intensive or individualized physical property is identified when "speed" is observed as the excretion of an individual in a confined area, capable of reaching someone one meter away after sneezing or coughing.
• This measure is represented in the unit of "meter per second", indicating its intensive nature.

At 30°C, glucose has a solubility of 1.25 g per gram of water. Given that the density of water at this temperature is 1 g/mL, the mass corresponding to 400 mL of water is also 400 g. Therefore, the concentration of the solution is calculated as 550 g divided by 400 g of water, which gives 1.375 g of glucose per gram of water. Since this concentration exceeds the solubility limit for glucose at this temperature, the solution can be classified as SATURATED.

Answer:

Empirical formula: Na3S4O9

Molecular formula: Na6S8O18

Explanation:

1) Upang makuha ang empirical formula, na siyang pinaka-simpleng buong bilang na ratio ng bawat elemento sa hindi kilalang compound na ito, una nating hahatiin ang dami ng bawat elemento sa kanilang sariling atomic mass.

Na= 23g/mol, S= 32g/mol, O=16g/mol

Na = 20.2/23 = 0.878mol

S = 37.6/32 = 1.175mol

O = 42.2/16 = 2.638mol

Susundan ito sa pamamagitan ng paghahati ng bawat bilang ng moles sa pinakamababang halaga (0.878mol)

Na = 0.878/0.878 = 1

S = 1.175/0.878 = 1.34

O = 2.638/0.878 = 3.004

Susunod, paramihin ang bawat decimal na bahagi ng 3 upang maging buo ang mga numero:

Na = 3

S= 4.02 ~ 4

O= 9.012 ~ 9

Kaya, ang pinaka-simpleng ratio ng Na, S, at O ay 3:4:9 kaya, ang empirical formula ay Na3S4O9

2) Para makuha ang molecular formula, gagamitin natin ang molecular weight ng bawat elemento sa empirical ratio, at ang kabuuang molar mass ng compound i.e.

Molar mass ng compound= 682.8g/mol

molecular weight ng Na3S4O9 = (3 x 23g/mol) + (4 x 32 g/mol) + (9 x 16 g/mol)

= 69 + 128 + 144

= 341 g/mol

Kaya, (Na3S4O9)n = 682.8g/mol

= (341)n = 682.8

n = 682.8/341

n~ 2

Kaya, molecular formula = (Na3S4O9)2

= Na6S8O18

From the table presented, C₆H₅O⁻ is the most frequently attached to the acidic hydrogen, due to its lower Ka (<span>acid dissociation constant). Chemical reaction: C₆H₅OH(aq) ⇄ C₆H₅O⁻(aq) + H⁺(aq). Ka = [C₆H₅O⁻] · [H⁺] / [C₆H₅OH]. The concentration of hydrogen ions in a phenol-water solution is very low.</span>

Be mindful of the integer next to each chemical in the equation, as these will be necessary for the stoichiometric calculations that follow:

Molar mass of <span>k[ag(cn)2]: 199 g/mol
Molar mass of Zn(Cn)2: 117.44 g/mol
Avogadro's number: 6.022</span>×10²³ molecules/mol

a.)

35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(6.022×10²³ molecules/mol) = 5.34×10²² molecules of Zn(Cn)₂

b.)

35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(117.44 g/mol) = 10.41 g of Zn(Cn)₂