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boyakko [2]
2 years ago
13

Calculate the molar mass of a gas at 388 torr and 45°c if 206 ng of gas occupies 0.206 ul. (

Chemistry
1 answer:
Dmitrij [34]2 years ago
6 0
To make our solution more systematic, let's convert all units that is consistent with the units of R which is 0.0821 L-atm/mol-K.

For pressure: 760 torr = 1 atm
388 torr * 1 atm/760 torr = 0.5105 atm

For volume, 1 μL = 10⁻⁶ L
0.206 μL * 10⁻⁶ L/1 μL = 2.06×10⁻⁷ L

For temperature,
T = 45 + 273 = 318 K

For mass, 1 ng = 10⁻⁹ g
206 ng * 10⁻⁹ g/1 ng = 2.06×10⁻⁷ g

Assuming ideal gas,

PV=nRT
(0.5105 atm)(2.06×10⁻⁷ L) = n(0.0821 L-atm/mol-K)(318 K)
n = 4×10⁻⁹ mol

Molar mass = Mass/n = 2.06×10⁻⁷ g/4×10⁻⁹ mol
<em>Molar mass = 51.14 g/mol</em>

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6. Elements that belong to the same group portray similar chemical properties. Therefore, the element in period 4 which is also in group 2 is <em>Calcium (Ca)</em>.

7. The elements that are striked through with the red slanting lines are the metalloids. All the elements to the left of the metalloids are metals. All the elements to the right are nonmetals. Bromine has a symbol of Br. Since At is a metalloid and located in the same group with Br, the <em>answer is Astatine (At).</em>

8. Tin has the chemical symbol of Sn. The nonmetal that is located in the same group is <em>Carbon (C)</em>.

9. All the elements in period 6 would have similar properties. The answers could be: <em>Phosphorus (P), Arsenic (As), Antimony (Sb) and Bismuth (Bi)</em>.

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5 0
2 years ago
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Which of the following is a reasonable ground-state electron configuration?
Veronika [31]

Answer is: 1s22s22p5.

1) Electron configuration 1s²1p⁶2d², this is not reasonable because 1p and 2d oritals do not exist.

2) Electron configuration 1s²2s⁴2p⁶, this is not reasonable because s orbitals only contain 2 electrons.

3) Electron configuration 1s²2s²2p⁵ is of an element fluorine. Fluorine (F) has atomic number 9, which means it has 9 protons and 9 electrons.

4) Electron configuration 1s²2s²2d⁶, this is not reasonable because 2d orbitals do not exist.

8 0
2 years ago
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What is the specific heat (J/g°C) of a metal object whose temperature increases by 3.0°C when 17.5 g of metal was heated with 38
Verizon [17]

Answer:

a. 0.73

Explanation:

Given data

  • Added heat (Q): 38.5 J
  • Change in the temperature (ΔT): 3.0°C
  • Mass of the metal (m): 17.5 g
  • Specific heat of the metal (c): ?

We can determine the specific heat of the metal using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 38.5 J / 17.5 g × 3.0°C

c = 0.73 J/g.°C

6 0
2 years ago
A solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What is the concentration of SO2 in ppm? in ppb?
Mnenie [13.5K]

Answer:

= 230 ppb

Explanation:

Considering that;

1ppm = 1mg/L  

Then;

0.00023g = 0.23mg  

Therefore;

0.00023 g/L = 0.23 mg/L

0.23 mg/L = 0.23 ppm

1 ppm = 100 ppb

Therefore;

0.23 ppm = 0.23 ×1000

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5 0
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consider the reaction between sulfite and a metal anion, X2-, to form the metal, X, and thiosulfate: 2 X2-(aq) + 2 SO32- + 3 H2O
Dafna11 [192]

Answer:

E_{red}^{0} for X is -1.20 V

Explanation:

Oxidation: 2\times[X^{2-}(aq.)-2e^{-}\rightarrow X(s)]

reduction: 2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)

---------------------------------------------------------------------------------------------------

overall:2X^{2-}(aq.)+2SO_{3}^{2-}(aq.)+3H_{2}O(l)\rightarrow 2X(s)+S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)

So, E_{cell}^{0}=E_{red}^{0}(SO_{3}^{2-}\mid S_{2}O_{3}^{2-})-E_{red}^{0}(X\mid X^{2-})

or, 0.63=-0.57-E_{red}^{0}(X\mid X^{2-})

or, E_{red}^{0}(X\mid X^{2-})= -1.20

So, E_{red}^{0} for X is -1.20 V

8 0
2 years ago
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