Answer:
13.4 (w/w)% of CaCl₂ in the mixture
Explanation:
All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.
To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.
<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>
0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻
<em>Moles CaCl₂:</em>
3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂
<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>
1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture
That means mass percent of CaCl₂ is:
0.207g CaCl₂ / 1.55g * 100 =
<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>
The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.
Answer:
The answer is True
Explanation:
The Ph of bady fluids have different ranges but it should be nearly constant to maintain the health of the human, The PH of the human blood is from 7.35 to 7.45 and the other blood fluids have different ranges . The PH shouldn't drop below 6.9 because it is so dangerous for human health and result coma. if there is high value of PH this indicates high numbers of OH- and low value of PH this indicates high numbers of H+
each organ has a certain PH value to do its function very well ,there is an organ needs acidic environment to work well and other organ needs an alkaline environment to do its work.
C) the number of the pricipal energy level containing the outer electrons increases
Answer:
Explanation:
1)The charge of one electron is given by
1 e = - 1.6 * 10-19 C
Then – 1 C = 1 e / ( 1.6 * 10-19 )
= 6.25 * 1018 e
So one-coulomb charge has 6.25 * 1018 electrons
2)Let q1 and q2 be two charges separated by a distance r
Then q1 = - 40 µC = - 40 * 10-6 C
And q2 = 108 µC = 108 * 10-6 C
