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2 years ago

Sulfurous acid, H2SO3, breaks down into water (H20) and sulfur dioxide (SO2).

Chemistry

1 answer:

erastovalidia [21]2 years ago

5 0

Answer:

1 atom of sulfur

Explanation:

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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

Item 4

Mariulka [41]

Answer:

Explanation:

8 0

3 years ago

Question 1 of 10What does the second quantum number (1) describe?O A. Which sublevel the electron is inB. Which energy level is

scoundrel [369]

Explanation:

The second quantum number also called the orbital quantum number describes the type of orbital or shape of it.

Answer: D. The specific orbital within a sublevel.

7 0

5 months ago

If a balloon containing 3000 L of gas at 39°C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature

ludmilkaskok [199]

$\tt =3000~L\times \dfrac{289}{312}\times \dfrac{99}{45.5}$

<h3>Further explanation</h3>

Given

3000 L of gas at 39°C and 99 kPa to 45.5 kPa and 16°C,

Required

the new volume

Solution

Combined with Boyle's law and Gay Lussac's law

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

T₁ = 39 + 273 = 312

T₂ = 16 + 273 = 289

Input the value :

V₂ = (P₁V₁.T₂)/(P₂.T₁)

V₂ = (99 x 3000 x 289)/(45.5 x 312)

or we can write it as:

V₂ = 3000 L x (289/312) x (99/45.5)

3 0

2 years ago

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

konstantin123 [22]

C3 H4 N1 ~~that’s what I think

8 0

2 years ago