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3 years ago

What is the radius of this circle if the circumference is 18 pi cm?

Chemistry

2 answers:

enot [183]3 years ago

5 0

Answer: 2.86 inches

Explanation: To find the radius, then, we insert 18 in for the circumference. So 18=2∏r. Solving for r gives 9/∏, or approximately 2.86 inches.

Jlenok [28]3 years ago

5 0

Answer:

2.86 inches

Explanation:

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What is one way to create a negative ion?

san4es73 [151]

Answer:

Burn beeswax candles. They clean the air, produce negative ions, and assist with allergy and asthma reduction. Run an indoor water fountain. The force or energy of falling water causes the neutral particles in air to split which frees electrons to attach to other air molecules resulting in negative ions.

Explanation:

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3 years ago

Uranium-235 has a half-life of 713 million years, while carbon-14 has a half-life of about 5,730 years. Which isotope would be m

nirvana33 [79]

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3 years ago

Read 2 more answers

Which of the following is an oxidation-reduction reaction? a. HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) b. Pb(C2H3O2)2(aq) + 2 NaCl

saw5 [17]

Answer: Option (d) is the correct answer.

Explanation:

A reaction in which there occurs change in oxidation state of reacting species is known as an oxidation-reduction reaction.

(a) $HCl(aq) + LiOH(aq) \rightarrow LiCl(aq) + H_{2}O(l)$

Will be written as:

$H^{+} + Cl^{-} + Li^{+} + OH^{-} \rightarrow Li^{+} + Cl^{-} + H^{+} + OH^{-}$

In this reaction, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(b) $Pb(C_{2}H_{3}O_{2})_{2}(aq) + 2NaCl(aq) \rightarrow PbCl_{2}(s) + 2 NaC_{2}H_{3}O_{2}(aq)$

Will be written as:

$Pb^{2+} + 2C_{2}H_{3}OO^{-} + 2Na^{+} + 2Cl^{-} \rightarrow Pb^{2+} + 2Cl^{-} + 2Na^{+} + 2C2H3OO^{-}$

Similarly here, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

Will be written as:

$Na^{+} + I^{-} + Ag^{2+} + NO^{2-}_{3} \rightarrow AgI(s) + Na^{+} + NO^{2-}_{3}(aq)$

Here, also there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(d) $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$

So, here there occurs change in oxidation state of Mg from 0 to +2 and oxidation state of H changes from +1 to 0. Hence, it is an oxidation-reduction reaction.

Thus, we can conclude that $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$ is an oxidation-reduction reaction.

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3 years ago

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AleksAgata [21]

Answer:

people can answer quicker withought writing out the equation.

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