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3 years ago

23) What is the volume of 2.454 x 1024 atoms of nitrogen gas at STP?

Chemistry

1 answer:

earnstyle [38]3 years ago

3 0

Answer:

d) V = 91.3 L

Explanation:

Given data:

Volume of nitrogen = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Number of atoms of nitrogen = 2.454×10²⁴ atoms

Solution:

First of all we will calculate the number of moles of nitrogen by using Avogadro number.

1 mole = 6.022×10²³ atoms

2.454×10²⁴ atoms × 1 mol / 6.022×10²³ atoms

0.407×10¹ mol

4.07 mol

Volume of nitrogen:

PV = nRT

1 atm × V = 4.07 mol ×0.0821 atm.L /mol.K ×273.15 K

V = 91.3 atm.L /1 atm

V = 91.3 L

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Otrada [13]

Answer:

A. 3 Significant Figures

B. 6 Significant Figures

C. 2 Significant Figures

D. 4 Significant Figures

E. 4 Significant Figures

F. 5 Significant Figures

Explanation:

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2 years ago

Write a nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90.

rusak2 [61]

Answer:

235/92U+10n→144/54Xe+90/38Sr+2/10n

Explanation:

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235/92U+10n→144/54Xe+90/38Sr+2/10n

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2 years ago

Name some acids and in what and how we used them

Ne4ueva [31]

Answer:

Acid Uses

Organic acids

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Explanation:

7 0

2 years ago

If 507 g FeCl2 were used up in the reaction FeCl2 + 2NaOH Fe(OH)2(s) + 2NaCl, how many grams of NaCl would be made?

Ket [755]

Number of moles of FeCl2 used = mass/ molar mass
Number of moles = 507/126.751 = 4.
If one mole of Fe reacts with two moles of sodium
Then 4 moles of Fe produces 8 moles of sodium.
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6 0

2 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

2 years ago