**Answer:**

4.8 %

**Explanation:**

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)

= 1010 g

% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %

Barium fluoride (BaF2) - Also known as Barium(II) fluoride - because it's a combination of two different kinds of ions (binary = two).

**Answer:**

The answer to your question is 3 moles of AlCl₃

**Explanation:**

**Process**

**1.- Write and balance the equation**

Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃

**2.- Determine the limiting reactant**

Theoretical proportion 1 mol Al(NO₃)₃ : 3 moles of NaCl

Experimental proportion 4 moles Al(NO₃)₃ : 9 moles NaCl

From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃ increases four times.

**3.- Determine the amount of AlCl₃ using proportions**

3 moles of NaCl --------------- 1 mol of AlCl₃

9 moles of NaCl ---------------- x

x = (9 x 1) / 3

x = 9 /3

**x = 3 moles**

The object has an overall positive charge.

**Answer:**

9 (1-2x²)

**Explanation:**

**The given expression is:**

30 - 9x²*2 - 21 - 4 + 4

**The first step is to compute the multiplication. This will give:**

30 - 18x² - 21 - 4 + 4

**Then, we will add like terms as follows:**

(30-21-4+4) - 18x²

= 9 - 18x²

**Finally, we can take the 9 as a common factor from both terms, this will give:**

9 (1-2x²)

Hope this helps :)