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podryga [215]
2 years ago
12

What volume of oxygen gas is needed to completely combust 0.202L of butane (C4H10) gas?

Chemistry
1 answer:
saveliy_v [14]2 years ago
4 0

Answer:

look at the file :)

Explanation:

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Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi
AlexFokin [52]

Explanation:

The given cell reaction is as follows.

       In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : In(s) \rightarrow In^{3+}(aq) + 3e^{-} ...... (1)

At cathode; Reduction-half reaction : Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s) ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

      2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}

      3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)

Net reaction: 2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

Thus, we can conclude that the overall cell reaction is as follows.

        2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

4 0
3 years ago
What reactions are involved in making plastic bags and what are the molecules that make up a plastic bag like?
denpristay [2]
Plastic bags are made up of plastic film, non-woven fabric, or plastic textile. The molecules that are present in plastic bags are polymers. Polymers are large molecules made up of repeating units called monomers.
I know this isn't much, but I hope it helps! :)
8 0
2 years ago
Describe how the height of the tides changes from Monday to Thursday.
adoni [48]
One of the results is that the moon is near the earth and the other one, the oceans tide. Even though the earth can hold any object within
ts proximity, the ocean is partly attracted due to its liquid property. At night, the ocean tends to be attracted to the moon by creating a bulge and assigning it as ‘high tide’. This is due to the strong gravitational pull of th moon to the earth.

I hope this helps!
This might be right..
5 0
3 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
3 years ago
10. What are the reactants in the process described below?
masha68 [24]

Answer:

don't know what to do

Explanation:

i don't know the answer

5 0
2 years ago
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