Answer:
7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.
Explanation:
Let the volume of 10% acid solution used to make the mixture = x L
So, the volume of 30% acid solution used to make the mixture = y L
Total volume of the mixture = <u>x + y = 30 L .................. (1)
</u>
For 10% acid solution:
C₁ = 10% , V₁ = x L
For 30% acid solution :
C₂ = 30% , V₂ = y L
For the resultant solution of sulfuric acid:
C₃ = 25% , V₃ = 30 L
Using
C₁V₁ + C₂V₂ = C₃V₃
10×x + 30×y = 25×30
So,
<u>x + 3y = 75 .................. (2)
</u>
Solving 1 and 2 we get,
<u>x = 7.5 L
</u>
<u>y = 22.5 L</u>
Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
INGREDIENTS:SUGAR, INVERT SUGAR, CORN SYRUP, MODIFIED CORN STARCH, CITRIC ACID, WHITE MINERAL OIL, NATURAL AND ARTIFICIAL FLAVOR, RED 40, AND CARNABUBA WAX.
Answer:
are those the answer choice
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
