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irinina [24]
2 years ago
11

Emperical formula of carbon dioxide

Chemistry
2 answers:
Vladimir79 [104]2 years ago
8 0

Empirical formula of Carbon dioxide :

  • \mathrm{CO_2}

Stells [14]2 years ago
6 0

CO2 is the emperical formula of carbon dioxide

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CH4 is insoluble in water 
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5. What is the number value of absolute zero? *<br> (
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Answer:

0

Explanation:

it is 0 I believe nothing

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What does kinetic energy depends on
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The kinetic energy of an object depends on two factors: mass(m) and velocity(v). The mass of an object can be measured in kilograms(kg) and velocity of the object in meters per second(m/s).
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Using lewis formulas, assign oxidation states to the atoms in these compounds. assume that sulfur is more electronegative than c
lilavasa [31]
Compound 1:     CS₂

Oxidation State of S  =  -2

Overall Charge on Molecule  = 0

So,
                             C + (-2)₂  =  0

                             C  -4  =  0
 
                             C  =  +4
Result: 
                         
   O.S of C  =  +4

                             
O.S of S  =  -2

Compound 2:     CH₃S-S-CH₃ or H₆C₂S₂

Oxidation State of H  =  +1

Oxidation State of S  =  -2

Overall Charge on Molecule  = 0
So,
                             (H)₆ + (C)₂ + (S)₂  =  0

                             (+1)₆ + (C)₂ + (-2)₂  =  0 

                             +6 + (C)₂ -4  =  0

                                 
   (C)₂  =  -6 + 4

                                     (C)₂  =  -2

                                       C  =  -1
Result: 
                         
    O.S of C  =  -1

                              
O.S of S  =  -2

                              
O.S of H  =  +1
3 0
3 years ago
Ammonium hydrogen sulfide decomposes according to the following reaction, for which Kp = 0.11 at 250°C: NH4HS(s) ⇌ H2S(g) + NH3(
Scrat [10]

Answer:

0,33atm

Explanation:

For the reaction:

NH₄HS(s) ⇌ H₂S(g) + NH₃(g)

kp is defined as:

kp = 0.11 = P(H₂S) P(NH₃) <em>(1)</em>

Where P(H₂S) and P(NH₃) are partial pressures of each compound.

In equilibrium, if in your system the only addition is of NH₄HS(s), the partial pressures and the concentration of each compound are:

NH₄HS: I - x

<em>-Where I is an initial concentration that is not relevant for the problem and x is the </em>NH₄HS<em> that reacts-</em>

H₂S(g): x

NH₃(g): x

Replacing in (1):

0.11 = X×X

0.11 = X²

<em>0.33 = X</em>

That means P(NH₃) is <em>0.33 atm</em>

<em></em>

I hope it helps!

5 0
3 years ago
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