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dimaraw [331]
3 years ago
8

How much did bottled water grow between 2015 and 2016?​

Chemistry
1 answer:
Solnce55 [7]3 years ago
7 0

Answer:

The International Bottled Water Association (IBWA), Alexandria, Va., and Beverage Marketing Corporation (BMC), New York, recently released 2015 bottled water statistics showing that Americans' consumption of bottled water increased by 7.9 percent and bottled water sales were up 8.9 percent from the previous year.

Explanation:

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Two solutions are created by mixing one solution containing lithium nitrate with one containing sodium phosphate.
babunello [35]

Answer:

Solution A that will form a precipitate with Ksp = 2.3 x 10−4

Explanation:

                                  Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

                                                     3S               S

Where S = Solubility(mole/lit) and Ksp = Solubility product

⇒ Ksp = (3S)³ x (S)

⇒ 27S⁴ = 2.3x10−4

⇒ S = 0.05 mol/lit

Concentration of Li₃PO₄ precipitate = 0.05

<u>Solution A </u>

0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole

0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole

                                     3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄

(Mole/Stoichiometry)    \frac{0.15}{3}                \frac{0.24}{1}

                                   = 0.05            = 0.24

Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.

So concentration of Li₃PO₄ is equal to 0.05.

                       

6 0
3 years ago
Consider the gas-phase equilibrium A ? B. In a series of experiments, different initial amounts of A and B are mixed together, a
AleksandrR [38]

Answer:

babbyy dee u wannna cookie

Explanation:

8 0
3 years ago
Consider the formation of nitrogen dioxide from nitric
Temka [501]

Answer:

9 L

Explanation:

According to the question , the given reaction is -

2NO(g) + O₂(g)------->2NO₂(g)

Since ,

At STP ,

One mole of a gas occupies the volume of 22.4 L.

Hence , as given in the question -

9 L of NO , i.e .

22.4 L = 1 mol

1 L = 1 / 22.4 mol

9 L = 1 / 22.4  * 9 L = 0.40 mol

From the chemical reaction ,

The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .

Hence ,  

2 moles of NO will produce 2 moles of NO₂.

Therefore ,

0.40 mol of NO will produce 0.40 mol of NO₂.

Hence , the volume of NO₂ can be calculated as -

1 mol = 22.4 L

0.40 mol = 0.40 * 22.4 L = 9 L

6 0
3 years ago
Arrange these elements based on their atomic radii. ga, f, s, as
mamaluj [8]

please have look at Periodic table , you will solve it yourself !

5 0
3 years ago
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.
Hitman42 [59]

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = 2 \times 0.475

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = 2 \times 0.025

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = pK_{a} + log \frac{Base}{Acid}  

              = 4.2 + log \frac{0.05}{0.045}

              = 4.245

For,  

         HCOOH + NaOH \rightarrow HCOONa + H_{2}O

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = pK_{a} + log \frac{Base}{Acid}  

          4.245 = 3.75 + log \frac{Base}{Acid}

      log \frac{Base}{Acid} = 0.5

    \frac{Base}{Acid} = 3.162

Now,

        \frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

                 

8 0
3 years ago
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