Answer:
d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.
Explanation:
From the reaction, 2Al(s) + 3Br2(l) → 2AlBr3(s)
2 moles of Al combines with 3 moles of Br to form 2 moles of AlBr3
Hence 1 mole of Br combines with 2/3 moles of Al
or 0.4 moles of Br combines with 2/3×0.4 moles of Al or 0.267 Moles of Al leaving
0.4 - 0.267 = 0.133 moles of Al remaining unreacted
The source of these two nitrogen atoms are ammonia (NH₃) from <span>nitrogen compounds (mostly metabolism of amino acids) through which excess nitrogen is eliminated from organisms. This process is called urea cycle, which extracted </span>nitrogenous wastes.
The liver<span> forms it by combining two </span>ammonia<span> molecules</span><span> with a </span>carbon dioxide<span> </span><span>molecule.</span><span />
Answer:
By visiting other households with cats.
Explanation:
This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.
Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles = 
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
