Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:
From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:
= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So, 
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
Answer:
V2 = 894.4mL
Explanation:
P1= 124.1, V1= 578mL, P2 = 80.2kPa, V2= ?
Applying Boyle's law
P1V1 = P2V2
Substitute and simplify
124.1*578=80.2*V2
V2= 894.4mL
To answer this problem, we need to count the electrons in the given configuration. The complete configuration is 1s2 2s2 2p6 3s2 3p6. There are 2+2+6+2+6 equal to 18 electrons. We find next the element with an atomic number of 18. That element is noble gas argon.
According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.
The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.
Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases
\frac{a}{b}=\frac{15}{35}=\frac{10}{y}
rearranging,
y=\frac{10\times 35}{15}=23.3
Thus, mass of b produced will be 23.3 g.
Answer:
C. BF3
Explanation:
The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.
