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3 years ago

Sofia goes on a hike on a trail that is 10 km long. She starts at 2:00pm and ends at

Chemistry

2 answers:

Butoxors [25]3 years ago

8 0

30 m/hrs (add me on discord if u have it Sipstic++#1460 )

Pavlova-9 [17]3 years ago

8 0

Answer:

100 m/hr S

Explanation:

Use formula Velocity = displacement/time

V=? d=300 t=3

V=300/3

V=100

Hope this helps!! :D

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Anna11 [10]

Answer:it’s C

Explanation:

A distant luminous object travels rapidly away from an observer.

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2 years ago

Identify the element using the Bohr model below.

Maurinko [17]

Nitrogen I believe . I need 20 characters.

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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.

6 0

3 years ago

For the reaction of hydrogen with iodine

fenix001 [56]

Answer:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Explanation:

Hello!

In this case, considering the given chemical reaction:

$H_2(g) + I_2(g) \rightarrow 2HI(g)$

Thus, by applying the law of rate proportions, we can write:

$\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}$

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Best regards!

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2 years ago

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Zolol [24]

The correct option is B. To increase the production of ammonia, you have to increase the pressure of the system. Increase in pressure will result in increased production of ammonia because this will drive the chemical reaction forward.

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3 years ago

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