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3 years ago

What is the average mass of a single argon atom in grams?

1 answer:

6 0

Best Answer:<span> </span><span>The molar mass of Ar = 39.948g/mol
1mol Ar = 6.022*10^23 atoms
Therefore 1 atom has mass = 39.948 /(6.022*10^23) = 6.63*10^-23g</span>

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12.00 moles of naclo3 will produce how many grams of O2? (Moles to Grams)

Sergeu [11.5K]

2NaClO₃ → 2NaCl + 3O₂

mole ratio of NaClO₃ to O₂ is 2 : 3

∴ if moles of NaClO₃ = 12 mol

then moles of O₂ = $\frac{12 mol * 3}{2}$

= 18 mol

Mass of O₂ = mol of O₂ × molar mass of O₂

= 18 mol × 16 g/mol

= 288 g

So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction. If you should have used another reaction then just follow the same steps and you'll get your answer.

4 0

3 years ago

The number of neutrons in the nucleus of an atom can be determined by

Masteriza [31]

Option no. 2 subtracting the atomic number from the mass number

Example: Zn - 65 atomic weight or mass/mass number
30 atomic number

65-30 = 35 number of neutrons

5 0

3 years ago

If an iron atom loses two electrons, the name of the iron ion is

adoni [48]

If iron atom looses 3 electrons, it forms Fe+3 (ferric) and if iron looses 2 electrons, it forms Fe+2(ferrous). Hope this helps!

5 0

3 years ago

Read 2 more answers

Methanol has a normal boiling point of 64.6C and a heat of vaporization of 35.2 kJ/mol. What is the vapor pressure (in Torr) of

DENIUS [597]

Answer:

vapor pressure of methanol at 12.0C = 75.09 torr

Explanation:

Using Clausius Clapeyron equation

, we have that

ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)

Given

At Normal boiling point,

Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm

Heat of vaporization = 35.2 kJ/mol

Changing to J/mol

=35.2 x 1000= 35200 J/mol

Temperature , T2 = 12.0oC = 12 + 273 = 285 K

Using gas constant, R = 8.314 J/mol.K

ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)

ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)

ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)

ln (P2/ 1 atm) = 4,233.822468 x-0.0005466866

ln (P2/ 1 atm)= -2.31457

P2 = e^⁻2.31457 x 1 atm

P2=0.098808atm

= 0.098808atm x760 = 75.09 torr

7 0

3 years ago

A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to

Snezhnost [94]

Explanation:

It is known that $pK_{a}$ of propionic acid = 4.87

And, initial concentration of propionic acid = $\frac{0.19}{1.20}$

= 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

= 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence, pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.216}{0.158}$

= 4.87 + log (1.36)

= 5.00

Therefore, when 0.02 mol NaOH is added then,

Moles of propionic acid = 0.19 - 0.02

= 0.17 mol

Hence, concentration of propionic acid = $\frac{0.17}{1.20 L}$

= 0.14 M

and, moles of sodium propionic acid = (0.26 + 0.02) mol

= 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.28 mol}{1.20 L}$

= 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.23}{0.14}$

= 4.87 + log (1.64)

= 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

Therefore, when 0.02 mol HI is added then,

Moles of propionic acid = 0.19 + 0.02

= 0.21 mol

Hence, concentration of propionic acid = $\frac{0.21}{1.20 L}$

= 0.175 M

and, moles of sodium propionic acid = (0.26 - 0.02) mol

= 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.24 mol}{1.20 L}$

= 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.2}{0.175}$

= 4.87 + log (0.114)

= 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0

3 years ago