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:
Guar gum
sodium nitrite
artificial food colorings
monosodium glutamate
etc
Answer is: hydrogen bonds.
Hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (H), covalently bound to a highly electronegative atom such as flourine (F), oxygen (O) and nitrogen (N) atoms.
According to the principle of base pairing hydrogen bonds could form between adenine and thymine (two hydrogen bonds between this nucleobases) and guanine and cytosine (three hydrogen bonds between this nucleobases).
Adenine and guanine are purine derivatives and thymine and cytosine are pyrimidine derivates.
Answer:
The answer is
<h2>11.25 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 45 g
volume = 4 mL
Substitute the values into the above formula and solve for the density
We have
We have the final answer as
<h3>11.25 g/mL</h3>
Hope this helps you
1.
V = 200 mL (volume)
c = 3 M = 3 mol/L (concentration)
First we convert mL to L:
200 mL = 0.2 L
Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol
Finally, we just use the molar mass of CaF2 to calculate the actual mass:
molar mass = 78 g/mol
The formula is: m = n × mm (mass = moles × molar mass)
m = 0.6 mol × 78 g/mol = 46.8 g
2.
For this question the steps are exactly like the first question.
V = 50mL = 0.05 L
c = 12 M = 12 mol/L
n = V × c = 0.05 L × 12 mol/L = 0.6 mol
molar mass (HCl) = 36.5 g/mol
m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.
3.
The steps for this question are the opposite way.
m(K2CO3) = 250 g
molar mass = 138 g/mol
n = m ÷ mm = 1.81 mol
c = 2 mol/L
V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL
