HELP!!! WILL MARK BRAINLIEST!!! picture attached below

Chemistry

1 answer:

Hoochie [10]2 years ago

5 0

Answer:

0.2 cm

Explanation:

looking at the ruler, you can see that the diameter is 0.4 cm. to get the radius, just divide by two, which gets 0.2 cm as the radius

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Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?

kifflom [539]

The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

5 0

3 years ago

Read 2 more answers

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:

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