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4 years ago

A sample of helium gas occupies 12.9 L at 315 K and 1.20 atm.

Chemistry

1 answer:

julsineya [31]4 years ago

5 0

Answer:

There are 0,6 moles of helium.

Explanation:

We use the ideal gas formula, we use the gas constant R = 0.082 l atm / K mol and we solve for n (number of moles) of the formula:

PV=nRT --> n= PV/RT

n= 1,20 atm x 12,9 L/ 0.082 l atm / K mol x 315 K= 0,599303135 moles

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A teacher makes the following statement. “Gasoline boils at a relatively low temperature (about 150°C). The kerosene is removed

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3 years ago

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Review the list of common titration errors. Determine whether each error would cause the calculation for moles of analyte to be

kakasveta [241]

Answer and Explanation:

A funnel is in the top of the buret and a beaker is positioned underneath the buret: This is correct and is necessary to fill the buret, but the funnel and the beaker has to be removed before the titration starts. The calculation for moles of analyte does not affect.

A solution is being poured from a bottle into the buret via the funnel: Using a funnel helps to fill the burette but it must be removed to filling the buret at 0.0 mL. In this case, the calculation for moles of analyte do not affect.

Adding titrant past the color change of the analyte solution: In this case, an excess of titrant is added, thus the calculation for moles of anality will be higher than it should be.

Recording the molarity of titrant as 0.1 M rather than its actual value of 0.01 M: In this case, the titrant is considered more concentrated than it is hence, the calculation for moles of anality will be higher than it should be.

Spilling some analyte out of the flask during the titration: The excess of titrant spilled out of the flask higher up the volume of titrant measured. Therefore, the calculation for moles of anality will be higher than it should be.

Starting the titration with air bubbles in the buret: The air inside the burette occupies measured volume, thus the volume of titrant measured will be higher than the real volume spilled in the flask. Hence the calculation for moles of anality will be higher than it should be.

Filling the buret above the 0.0 mL volume mark: Some volume of titrant will be spilled inside the flask but will no be measured since the buret measures the titrant below the 0.0mL mark, thus the calculation for moles of anality will be lower than it should be.

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3 years ago

Gases Solids compress more easily than liquids or gases. True False

SSSSS [86.1K]

Answer:

False

Explanation:

Gasses and have more space between molecules

4 0

4 years ago

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SYNTHESIS OFCARBONATECTIONLABORATORY SIMULATIONLab Data- X99.00.10CollectedVolume sodium carbonate (mL)Molarity sodium carbonate

evablogger [386]

Answer:

$\begin{gathered} \text{Limiting Reagent = Sodium Carbonate} \\ \text{Percent Yield = 98\%} \end{gathered}$

Explanation:

The chemical reaction talks about the synthesis of calcium carbonate

It is from the reaction between sodium carbonate and calcium chloride

Let us write the equation of reaction as follows:

$Na_2CO_{3(aq)}+CaCl_{2(aq)}\text{ }\rightarrow2NaCl_{(s)\text{ }}+CaCO_{3(aq)}$

Firstly, we want to get the expected mass of calcium carbonate

This speaks about getting the theoretical yield based on the equation of reaction

From the data collected, 90 ml of 0.20 M (mol/L) of sodium carbonate gave calcium carbonate

We need to get the actual number of moles of sodium carbonate that reacted

We can get this by multiplying the volume by the molarity (kindly note that we have to convert the volume to Liters by dividing by 1000)

Thus, we have it as:

$\frac{90}{1000}\times\text{ 0.1 = 0.009 moles}$

Hence, we see that 0.009 moles of sodium carbonate reacted theoretically

Since 1 mole of sodium carbonate gave 1 mole calcium carbonate, it is expected that 0.009 mole of sodium carbonate will give 0.009mole of calcium carbonate

What we have to do now is to get the theoretical grams of calcium carbonate produced

That would be the product of the number of moles of calcium carbonate and its molar mass

The molar mass of calcium carbonate is 100 g/mol

The theoretical yield (expected mass) is thus: