Answer:
a. the mole fraction of CO in the mixture of CO and O2.
mole fraction = moles of CO/ Total moles of the mixture
Mole fraction of CO = 10/(10+12.5)=0.444
b. Reaction - CO(g)+½O2(g)→CO2(g)
Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2
So given,
At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.
3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2
This means that unused mols are : 7mols of CO and 11mols of O2
Total product mixture = 3 + 7 + 11 = 21mols
mole fraction of CO = 7/21 = 0.33
It shouldn't have any lone pairs since it is a tetrahedral structure. Ge has 4 valence electrons. Each H has 1 valence electron. Therefore, each H valence electron will pair with each valence electron on Ge.
Answer:
-68.4 kJ
Explanation:
<u>The standard enthalpy of vaporization = 23.3 kJ/mol</u>
<u>which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).</u>
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
<u>Thus, Q = -23.3 kJ/mol</u>
<u>Where negative sign signifies release of heat</u>
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
<u>Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.</u>
Answer:
C4H8
Explanation:
First find the molar mass of CH2;
2(1.01) + 1(12.01) = 14.03g
Now divide the molar mass of the compound by the molar mass of CH2;
56g/14.03g = 3.9914 Round to nearest whole number = 4
Multiply CH2 by 4 to get the molecular formula;
CH2* 4 = C4H8
Answer:
Conductors
Explanation:
Metals that are conductors let electric currents flow freely. Insulators have a resistance of a charge to flow through them.
