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3 years ago

Using the information from STP or SATP conditions determine the value of the ideal gas constant.

Chemistry

1 answer:

Ugo [173]3 years ago

8 0

Answer:

0.0821 atm.L/Kmol

Explanation:

At stp, the values temperature, pressure and volume is given below:

Pressure (P) = 1 atm

Temperature (T) = 273 K

Volume (V) = 22.4 L

At stp, 1 mole of a gas occupy 22.4L.

Number of mole (n) = 1 mole

Gas constant (R) =?

The ideal gas equation is given below:

PV = nRT.

With the above equation, the gas constant R can be obtained as follow:

1 atm x 22.4L = 1 mole x R x 273K

Divide both side by (1 mole x 273 K)

R = (1 atm x 22.4L) / (1 mole x 273 K)

R = 0.0821 atm.L/Kmol

Therefore, the gas constant is 0.0821 atm.L/Kmol

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Given: There are 39.95 grams of Argon (39.95 g/1 mole) and one mole has a volume of 22.4 Liters (1 mole/22.4 L). What is the vol

marta [7]

Answer:

$V_2=19.23L$

Explanation:

Hello,

In this case, by using the Avogadro's law which allows us to understand the volume-moles behavior as a directly proportional relationship:

$\frac{V_2}{n_2} =\frac{V_1}{n_1}$

We can compute the volume of 34.3 g of argon by representing it in mole as shown below:

$n_1=1 mol\\\\n_2=34.3g*\frac{1mol}{39.95g} =0.859mol$

Thus, we find:

$V_2=\frac{V_1*n_2}{n_1}=\frac{22.4L*0.859mol}{1mol} \\\\V_2=19.23L$

Best regards.

6 0

3 years ago

For the reaction 2Na + Cl2 → 2NaCl, how many grams of chlorine gas are required to react completely with 57.5 mol of sodium?

Step2247 [10]

Answer:

1019.27 g.

Explanation:

For the balanced reaction:

It is clear 2 moles of Na with 1 mole of Cl₂ to produce 2 moles NaCl.

Firstly, we need to calculate the no. of moles of Cl₂ is needed to react with 57.5 mol Na:

2 moles of Na need → 1 mol of Cl₂, from the stichiometry.

57.5 moles of Na need → ??? mol of Cl₂.

<em>∴ The no. of moles of Cl₂ is needed to react with 57.5 mol Na =</em> (1 mol)(57.5 mol)/((2 mol) <em>= 28.75 mol.</em>

<em>∴ the mass of Cl₂ is needed to react with 57.5 mol Na = (no. of moles of Cl₂)(molar mass of Cl₂) =</em> (28.75 mol)(35.453 g/mol) <em>= 1019.27 g.</em>

4 0

3 years ago

What characteristic of an element’s atoms always determines the element’s identity?

Anna11 [10]

Answer:

hporntue dhdjehwgs r. rvegdyfuee

Explanation:

ehehrhrhrhrhrhrhr. dvdhdhrhrhehehr f fbdhehrgdgdhehd dbdh

3 0

4 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Balance the following equation<br> mg(s)+ HCl H2)g)+MgCl2

Marat540 [252]

<h3>Given equation:-</h3>

$\\ \sf\longmapsto Mg+HCl\longrightarrow MgCl_2+H_2$

<h3>BALANCED EQUATION:-</h3>

$\\ \sf\longmapsto Mg+2HCl\longrightarrow MgCl_2+H_2$

6 0

3 years ago