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Juliette [100K]
3 years ago
15

Name the compound O2Cl6

Chemistry
1 answer:
MrMuchimi3 years ago
7 0

Answer:

dioxide hexachloride

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Calculate the ΔHrxn for the following
Vesnalui [34]

Answer:

2 NO (g) → N2 (g) + O2 (g)

2 NOCl (g) → 2 NO (g) + Cl2 (g)

____________________________

2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)

ΔH = [90.3 kJ x 2 x -1] +  [-38.6 kJ x -1 x 2] = -103.4 kJ

The ΔH for the reaction is -103.4 kJ

6 0
3 years ago
How much heat will be released when 10.0 g of hydrogen peroxide decomposes according to the following reaction: 2H2O2(l) 2H2O(l)
kramer
C. 28 KJ

AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2

0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
5 0
3 years ago
Why can’t you see the stars very well at night like you could 50 years ago
Elenna [48]

Answer:Artificial light from cities has created a permanent "skyglow" at night, obscuring our view of the stars. Here's their map of artificial sky brightness in North America, represented as a ratio of "natural" nighttime sky brightness. In the black areas, the natural night sky is still (mostly) visible.

Explanation:

6 0
2 years ago
3. What noble gas would be part of the electron configuration notation for Mn?
shusha [124]

Answer:

Argon {Ar}

Explanation:

The noble gas used for a condensed electron configuration is the one before the element which you are configuring.

In this case, the element (Mn) is manganese

The noble gas that is before this element is Argon which is the row above it

so your configuration would be {Ar} 3d^5 4s^2

7 0
3 years ago
Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °
Mila [183]
This problem is to use the Claussius-Clapeyron Equation, which is:

ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]

Where p2 and p1 and vapor pressure at estates 2 and 1

ΔH is the enthalpy of vaporization

R is the universal constant of gases = 8.314 J / mol*K

T2 and T1 are the temperatures at the estates 2 and 1.

The  normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa

Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol 

=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]

=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x =  0.000157 + 1/330.95 = 0.003179

=> x = 314.6 K => 314.6 - 273.15 = 41.5°C

Answer: 41.5 °C 
3 0
3 years ago
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