Astroid is the answer to the problem
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.
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Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Answer:
Detail is given below
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.
In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm
In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N
In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.
