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steposvetlana [31]
4 years ago
5

a gas that exerts a pressure of 215 torr in a container with a volume of 51.0 mL will exert a pressure of ? torr when transferre

d to a container with a volume of 18.5L
Chemistry
1 answer:
zhannawk [14.2K]4 years ago
3 0
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

P1V1 = P2V2,

Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:

(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr

The final pressure exerted by the gas would be 0.593 torr.

Hope this helps!
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1. For every 5.00 mL of milk of magnesia there are 400. mg of magnesium hydroxide. How many mL of milk of magnesia do we need to
forsale [732]

Answer:

1. 6.50mL

2. pH = 13.097

Explanation:

1. The neutralization of HCl with Al(OH)₃ is:

3HCl + Al(OH)₃ → Al(Cl)₃ + 3H₂O

40.0mL ≡ 0.0400L of 0.500M HCl are:

0.0400L × (0.500mol / L) = <em><u>0.0200 moles of HCl</u></em>

Based on the neutralization, 3 moles of HCl react with 1 mole of Al(OH)₃, thus, the moles of Al(OH)₃ that react with 0.0200 mol of HCl are:

0.0200mol HCl × (1mol Al(OH)₃ / 3mol HCl) = <em><u>0.00667 moles of Al(OH)₃</u></em>. In grams:

0.00667 moles Al(OH)₃ × (78g / 1mol) = 0.520g of Al(OH)₃ ≡ 520mg

As 5.00mL of milk magnesia contain 400mg of Al(OH)₃, the mL of milk magnesia required for a complete reaction are:

520 mg Al(OH)₃ × (5.00mL / 400mg) =<em> 6.50mL</em>

<em />

2. The reaction of lithium hydroxide (LiOH) with perchloric acid (HClO₄) is:

HClO₄ + LiOH → LiClO₄ + H₂O

<em>The reaction is 1:1.</em>

Moles of LiOH and HClO₄ are:

LiOH: 0.0500L × (0.350mol / L) = <em>0.0175 moles of LiOH</em>

HClO₄: 0.0300L × (0.250mol / L) = <em>0.0075 moles of HClO₄</em>

Assuming the reaction goes to completion, moles of LiOH that remains are:

0.0175 mol - 0.0075 mol = 0.0100 moles of LiOH. The total volume is 80.0mL, 0.0800L. Thus, molarity of LiOH is:

0.0100 mol / 0.0800L = <em>0.125 M of LiOH</em>

It is possible to obtain the pOH of the solution thus:

pOH = -log (OH) = -log 0.125M = 0.903

As pH = 14- pOH,

pH = 14 - 0.903 = <em>13.097</em>

I hope it helps!

3 0
3 years ago
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