Answer:
660kcal
Explanation:
The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.
We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:
3.3L * (1000mL / L) * (5g/100mL)= 165 g.
If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.
E=mc (square) E= mass times capacity squared
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Answer:
Option b. Decomposition
Followed by a reduction process using charcoal
Explanation:
Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:
2Pb(NO3)2 —> 2PbO + 4NO2 + O2
The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:
2PbO + C —> Pb + CO2
Uhh add a picture so I can help
