The mass of required will be 0.6075 g.
Calculation of mass:
The reaction:
Volume of = 75 mL
Concentration of = 0.100 M
Step 1:
Molarity is moles of solute in 1000 mL of the solution.
Moles of = 0.0075
Step 2:
From the above equation,
2 moles of is precipitated by 1 mole of
So, 0.0075 moles will be precipitated by-
moles of
= 0.00375 moles of
Step 3:
Molar mass of = 162 g/mol
So, Mass of required will be 0.6075 g.
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Answer:
Metallic.
Explanation:
Hello,
In this case, when nickel and tin bond, the difference in their electronegativities results:
Such difference, in addition to the fact that nickel is a transition metal and tin a metal, will suggest that the predominant type of bond for this substance is metallic as the attractive force between the conduction of electrons and positively charged metal ions is present.
Best regards.
Answer:
The answer to your question is 0.6 M
Explanation:
Data
Molarity = ?
Volume = 450 ml
mass of NaCl = 15 g
Process
1.- Calculate the molar mass of NaCl
NaCl = 23 + 35.5 = 58.5 g
2.- Calculate the number of moles of NaCl
58.5 g of NaCl ------------ 1mol
15 g of NaCl ------------ x
x = (15 x 1)/58.5
x = 0.26 moles
3.- Calculate the molarity
Molarity = moles/volume
-Substitutiion
Molarity = 0.26/0.45l
-Result
Molarity = 0.6 M
Hydrogen gas and oxygen gas react to form liquid water according to the following equation:
2H₂ + O₂ → 2H₂O
a. Converting our given masses of each gas to moles, we have:
(25 g H2)/(2 × 1.008 g/mol) = 12.4 mol H2; and
(25 g O2)/(2 × 15.999 g/mol) = 0.781 mol O2.
From the equation, two moles of H2 react with every one mole of O2. To fully react with 12.4 moles of H2, as we have here, one would need 6.2 moles of O2, which is far more than what we're actually given. Thus, the oxygen is our limiting reactant, and as such it will be the first reactant to run out.
b. Since O2 is our limiting reactant, we use it for determining how much product, in this case, H2O, is produced. From the equation, there is a 1:1 molar ratio between O2 and H2O. Thus, the number of moles of H2O produced will be the same as the number of moles of O2 that react: 0.781 moles of H2O. The mass of water produced would be (0.781 mol H2O)(18.015 g/mol) ≈ 14 grams of water (the answer is given to two significant figures).
c. Since the hydrogen reacts with the oxygen in a 2:1 ratio, twice the number of moles of oxygen in hydrogen is consumed: 0.781 mol O2 × 2 = 1.562 mol H2. Since we began with 12.4 moles of H2, the remaining amount of excess H2 would be 12.4 - 1.562 = 10.838 mol H2. The mass of the excess hydrogen reactant would thus be (10.838 mol H2)(2 × 1.008 g/mol) ≈ 22 grams of hydrogen gas (the answer is given to two significant figures).