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3 years ago

5

Where do the OH- ions go as additional acid is added?

2 answers:

qwelly [4]3 years ago

8 0

The result when the OH- ions go as additional acid is added is that t<span>hey remain in solution but change concentration. </span>This is because when you add more acid to hydroxide, it will change its concentration due to the acidity added.

stiks02 [169]3 years ago

8 0

Answer:

A: They evaporate

Explanation:

Where do the OH- ions go as additional acid is added? They evaporate. They combine with excess H+ ions to form more water. They remain in the solution at the same concentration.

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Does sodium satisfy the ostet rule?

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3 years ago

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3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

A sample of a vapor occupies a volume of 500 mL at 65°C. If pressure remains constant, what is the volume of the gas at standard

DedPeter [7]

Answer:

403 mL

Explanation:

First, I will assume that the mole is 1, because you are not specifing this.

Now, with the innitial data, we need to get the pressure:

T = 65+273 = 338 K

V = 500 / 1000 = 0.5 L

Now if:

PV = nRT

Then:

P = nRT/V and V = nRT/P

Let's calculate the P:

P = 1 * 0.082 * 338 / 0.5 = 55.432 atm

The standard temperature is 0° C or 273 K so, the volume is:

V = 1 * 0.082 * 273 / 55.432

V = 0.40384 L or simply 403.84 mL

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3 years ago