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masya89 [10]
3 years ago
7

You are making lemonade and the calls for 6 cups Lemon juice (L), 3 Cups of sugar (S) and 5 cups of water (W) to make 12 Cups of

Lemonade (LA): your formula is:
6 L + 3 S + 5 W = 12 LA
You are given 9 cups of lemon juice (L), 8 cups of sugar (S) and unlimited water (W).
How many cups of lemonade can you make?
1. 9
2. 12
3. 18
4. 32
Chemistry
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

18 Cups

Submitted the assignment

Explanation:

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2 years ago
If aqueous solution of lead (ll) nitrate and sodium sulfate, which insoluble precipitate is formed
forsale [732]

Answer:

Lead(II) sulfate

Explanation:

This looks like a double displacement reaction, in which the cations change partners with the anions.

The possible products are

Pb(NO₃)₂ (aq)+ Na₂SO₄(aq) ⟶PbSO₄(?) + 2NaNO₃(?)

To predict the product, we must use the solubility rules. Two important ones for this question are:

  1. Salts containing Group 1 elements are soluble.
  2. Most sulfates are soluble, but PbSO₄ is an important exception.

Thus, NaNO₃ is soluble and PbSO₄ is the precipitate.

7 0
4 years ago
If the concentration of a KCl solution is 16.0% (m/v), then the mass of KCl in 26.0 mL of solution is ________.
melomori [17]

Answer:

The correct answer is 4.16 grams.

Explanation:

Based on the given information, the concentration of KCl solution is 16 % m/v, which means that 100 ml of the solution will contain 16 grams of KCl.

The molarity of the solution can be determined by using the formula,

M = weight/molecular mass × 1000/Volume

The molecular mass of KCl is 74.6 grams per mole.

M = 16/74.6 × 1000/100

M = 16/74.6

M = 2.14 M

Now the weight of KCl present in the solution of 26 ml will be,

2.14 = Wt./74.6 × 1000 /26

Wt. = 4.16 grams

3 0
3 years ago
A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka
raketka [301]

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

  • HA ⇄ H⁺ + A⁻
  • Ka = [H⁺][A⁻]/[HA]

We <u>calculate [H⁺] from the pH</u>:

  • pH = -log[H⁺]
  • [H⁺] = 10^{-pH}
  • [H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

  • [HA] * 0.66/100 = [H⁺]

We <u>calculate [HA]</u>:

  • [HA] = 0.138 M

Finally we <u>calculate the Ka</u>:

  • Ka = \frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]} = 6.02x10⁻⁶
3 0
3 years ago
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