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.
Answer:
V₂ = 1070 mL or 1.07 L
Solution:
Data Given;
P₁ = 1170 mmHg
V₁ = 915 mL
T₁ = 24 °C + 273 K = 297 K
P₂ = 842 mmHg
V₂ = ?
T₂ = - 23 °C + 273 K = 250 K
According to Ideal gas equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / P₂ T₁
Putting Values,
V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)
V₂ = 1070 mL or 1.07 L
An electron in the 3s orbital. The order of electron orbital energy levels starting from lowest to highest is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
Answer:
All ions must a overall charge
Explanation:
