B. sublimation is the answer.
For example, monosaccharides:
C₆H₁₂O₆
aldohexoses
ketohexoses
C₃H₆O₃
aldotrioses
ketotrioses
and many others
Answer:
81.5g/mol
Explanation:
Molar mass is the ratio between mass of a substance (In this case, 0.207g) and moles presents in this mass.
To solve this question we must find the moles of the gas in order to obtain the molar mass using:
PV = nRT
PV / RT = n
<em>Where P is pressure = 0.980atm</em>
<em>V is volume in Liters = 0.0725L</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is absolute temperature = 68°C + 273.15 = 341.15K</em>
<em />
0.980atm*0.0725L / 0.082atmL/molK*341.15K = n
2.54x10⁻³ moles = n
Thus, the molar mass of the gas is:
0.207g / 2.54x10⁻³ moles
<h3>81.5g/mol</h3>
2NaHCO3 -> Na2CO3 + H2O + CO2
<span>2.765g NaHCO3/MM = moles NaHCO3 </span>
<span>moles NaHCO3 x (1 mole Na2CO3 / 2 moles NaHCO3) x MM Na2CO3 = theoretical yield of Na2CO3 </span>
<span>Percent yield is simply the actual yield/theoretical yield (x100 to put it into percentage).
MM = Molar mass (grams of substance per mol)</span>
I cause the water to het up, when it do it become a gas.
