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3 years ago

PLEASE HELP ME BUT FAST!

Chemistry

2 answers:

Elina [12.6K]3 years ago

7 0

Hey, i add an answer so you can give the other person brainlest :)

Lisa [10]3 years ago

7 0

G - goes in the past showing the different faces on the coins
O - on/ in the 1900's they were made
L - liquid when past its boiling temperature
D - damage resistant
sorry that these are bad i was struggling with these

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The heat exchange in chemical reactions is due to a change in:

Paha777 [63]

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Explanation:

This would be due to enthalpy. This is because that this where heat exchange happens in chemical reactions.

So, a positive enthalpy means that the reaction is endothermic (heat entering) and if enthalpy is negative this will be a exothermic (heat released)

Another example would be the use of atoms. This example if temperature is the factor, this means that at low temp there is less kinetic energy so low temperature. However, an increase in energy means that there is more kinetic energy in the atoms which means that more collisions occurring.

Hope this helps !!!!

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3 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

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