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alexandr1967 [171]
2 years ago
15

What is the molar mass of 56.75 g of gas exerting a pressure of 2.87 atm on the walls of a 5.29 l container at 230 k?

Chemistry
1 answer:
laiz [17]2 years ago
8 0
We first need to find the number of moles of gas in the container 
PV = nRT
where;
P - pressure - 2.87 atm x 101 325 Pa/atm = 290 802.75 Pa
V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation 
290 802.75 Pa x  5.29 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 230 K
n = 0.804 mol
the molar mass = mass present / number of moles
molar mass of gas = 56.75 g / 0.804 mol 
therefore molar mass is 70.6 g/mol 
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Four balloons, each with a mass of 10.0 g, are inflated to a volume of 20.0 L, each with a different gas: helium, neon, carbon m
weeeeeb [17]
On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
6 0
2 years ago
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If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/
riadik2000 [5.3K]

<u>Answer:</u> The energy of the complex is 2.72\times 10^2kJ

<u>Explanation:</u>

To calculate the energy of the complex, we use the equation given by Planck which is:

\Delta E=\frac{N_Ahc}{\lambda}

where,

\lambda = Wavelength of the complex = 440nm=4.40\times 10^{-7}m    (Conversion factor:  1m=10^9nm )

h = Planck's constant = 6.624\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

N_A = Avogadro's number = 6.022\times 10^{23}

\Delta E = energy of the complex

Putting values in above equation, we get:

\Delta E=\frac{6.022\times 10^{23}\times 6.624\times 10^{-34}\times 3\times 10^8}{4.40\times 10^{-7}}\\\\\Delta E=2.72\times 10^{5}J=2.72\times 10^2kJ

Conversion factor used:  1 kJ = 1000 J

Hence, the energy of the complex is 2.72\times 10^2kJ

5 0
2 years ago
A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

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Step 2: Density of the gold bar

Density = Mass ÷ volume

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Thus, the density of the gold bar is 19.3 g/cm³

3 0
2 years ago
Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative
koban [17]

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

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Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

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percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

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6.941 =  45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

The atomic mass of second isotope is 7.016

3 0
2 years ago
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