All categories

3 years ago

"sodium hydroxide is a strong base because it absorbs more" ____________ ions than weak bases.

Chemistry

2 answers:

pentagon [3]3 years ago

7 0

Answer:

Hydrogen ions

Explanation:

Neutralization reaction is actually a reaction between hydroxide ions and hydrogen ions to form water.

H^+(aq) + OH^-(aq) ----------> H2O(l)

A strong base can absorb more hydrogen ions than a weak base to form more water molecules as shown above.

That is the difference between a weak base and a strong base.

Hunter-Best [27]3 years ago

4 0

The answer is: absorbs more H (protons) ions.

- The Sodium hydroxide NaOH ionizes completely when dissolved in water.
- For every mole of sodium hydroxide that you dissolve you get 1 mole of hydroxide anions.

You might be interested in

Earth's atmosphere, which traps heat and radiation from the sun, helps to keep our temperature somewhat stable, even when we are

Elis [28]

Answer:

It is true

Explanation:

6 0

3 years ago

Read 2 more answers

17 points if you say hi to claim points ur gay.

nikitadnepr [17]

Answer:

Explanation:

Boiling point

6 0

3 years ago

Read 2 more answers

Can someone help me with problem 21.105? Many thanks!

Reika [66]

The correct answer is 1 to the 3rd power

3 0

2 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

A ball is rolled along and off a table. Which of the following acts on the ball after it leaves the

alexira [117]

Both the force of the earth’s gravity on the ball and the force the ball got from being rolled off

6 0

2 years ago