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charle [14.2K]
3 years ago
12

If 50 milliliters of 0.50 M HCl is used to completely neutralize 25 milliliters of KOH solution, what is the molarity of the bas

e?
Chemistry
1 answer:
NemiM [27]3 years ago
3 0

Answer:

0.25 M

Explanation:

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Please help! Thanks
trapecia [35]

Th actual yield of the reaction is 24.86 g

We'll begin by calculating the theoretical yield of the reaction.

2Na + Cl₂ → 2NaCl

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl from the balanced = 2 × 58.5 = 117 g

From the balanced equation above,

46 g of Na reacted to produce 117 g of NaCl.

Therefore,

11.5 g of Na will react to produce = (11.5 × 117) / 46 = 29.25 g of NaCl.

Thus, the theoretical yield of NaCl is 29.25 g.

Finally, we shall determine the actual yield of NaCl.

  • Percentage yield = 85%
  • Theoretical yield = 29.25 g
  • Actual yield =?

Actual yield = Percent yield × Theoretical yield

Actual yield = 85% × 29.25

Actual yield = 0.85 × 29.25 g

Actual yield = 24.86 g

Learn more about stoichiometry: brainly.com/question/25899385

7 0
2 years ago
84. Predict how many electrons each element will most likely
Kruka [31]

Answer:

I will gain 1 electron

Ba will lose 2 electrons

Cs will lose 1 electron

Se will lose 2 electrons

Explanation:

5 0
3 years ago
Calculate the hydrogen ion concentration in a 3.4 x 10-3 M<br> solution of KOH.
notsponge [240]

Explanation:

KOH→K {}^{ + }  + OH {}^{ - }  \\ [OH {}^{ - } ] = 3.4 \times  {10}^{ - 3}  \\ k _{w} = [OH {}^{ - } ] [H {}^{  + } ]  \\ 1 \times  {10}^{ - 14}  = 3.4 \times  {10}^{ - 3}  \times [H {}^{  +  } ]  \\ [H {}^{  +  } ]  = 2.94 \times  {10}^{ - 12}  \: M

7 0
3 years ago
Help me please!!!!!!
VashaNatasha [74]

3 AgN03 + 1 Al ---> 3 Ag + 1 Al(NO3)3

5 0
3 years ago
A 17.3 g sample of Caso, is found to contain 5.09 g of Ca and 8.13 g of O. Find the mass of sulfur in a sample of CaSo, with a
hammer [34]

Answer:

4.08g

Explanation:

17.3=(5.09+8.13+S)

4 0
3 years ago
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