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myrzilka [38]
2 years ago
14

. If one mole each of CH4, NH3, H2S, and CO2 is added to 1 liter of water in a flask (1 liter of water = 55.5 moles of H2O), how

many moles of hydrogen, carbon, oxygen, nitrogen, and sulfur are in the flask? Round your answers to the nearest whole number.
How many moles of Hydrogen, Carbon, Oxygen, Nitrogen, and sulfur will there be?
Chemistry
1 answer:
denpristay [2]2 years ago
6 0
1 litre of water is = 55.5 moles of water.
water is H2O
so, in water:
moles of oxygen = 55.5
moles of hydrogen = 2 x 55.5 = 111

Now, 1 mole each of <span>CH4, NH3, H2S, and CO2 are added:
For CH4: 
moles of C = 1
moles of H = 4 x 1 = 4

For NH3:
moles of N = 1
moles of H = 3 x 1 = 3

For H2S:
moles of H = 2 x 1 = 2
</span>moles of S = 1
<span>
For CO2:
</span>moles of C = 1
moles of  = 2 x 1 = 2
<span>
Now, add the total moles of each atom:
Hydrogen = 111 + 4 + 3 +1 = 119 moles
Oxygen = 55.5 + 2 = 57.5
Carbon = 1+1 = 2
Sulfur = 1
nitrogen = 1

</span>
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Answer:

Look on the picture.

Explanation:

He could find only 2 isomers of n-hexane alkenes for this reaction. Other two could be marked from other direction.

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The standard heats of combustion (δh∘) per mole of 1,3-butadiene, c4h6(g); butane, c4h10(g); and h2(g) are −2540.2,−2877.6, and
ryzh [129]

solution:

Hydration is the addition of water; hydrogenation is the addition of hydrogen.  

desire rxn: _C4H6(g) + 2 H2(g)-----> C4H10(g)___dHhy = ??  

knowns:  

__________C4H6 + 11/2 O2 --------> 4CO2 + 3H2O______dHox = -2540.2 kJ/mole  

__________4CO2 + 5H2O -----------> C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole  

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Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem


5 0
2 years ago
Which statement correctly describes the location and charge of the electrons in an atom?
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Answer:

D

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3 0
1 year ago
Draw a lewis structure for so2 in which all atoms obey the octet rule. show formal charges. do not consider ringed structures.
Aleonysh [2.5K]
Formal charge = valence electron - bonds - dots.
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6 0
1 year ago
Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with
stepan [7]

Answer:

1) Net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2) 0.765 M is  the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq) ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq) ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq) ..[3]

H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)

From one:[1] ,[2] and [3]:

2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)

Cancelling common ions on both sides to get net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2CO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL

Putting values in above equation, we get:

M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M

0.765 M is  the molarity of the carbonic acid solution.

6 0
2 years ago
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