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sdas [7]
3 years ago
8

I am a subatomic particle that is positively charged what am I?

Chemistry
1 answer:
lions [1.4K]3 years ago
6 0

<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em><u>Proton</u></em><em><u>.</u></em>

<em><u>Additional</u></em><em><u> </u></em><em><u>Information</u></em><em><u>:</u></em>

<em><u>There</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>three</u></em><em><u> </u></em><em><u>types</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>subatomic</u></em><em><u> </u></em><em><u>particles</u></em><em><u>.</u></em><em><u> </u></em><em><u>They</u></em><em><u> </u></em><em><u>are</u></em><em><u>:</u></em>

  1. <em><u>Proton</u></em>
  2. <em><u>Electron</u></em>
  3. <em><u>Neutron</u></em>

<em><u>Proton</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>positively</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particle</u></em><em><u>,</u></em><em><u>Electron</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>negatively</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particle</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>Neutron</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>charge</u></em><em><u> </u></em><em><u>less</u></em><em><u>.</u></em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>be</u></em><em><u> </u></em><em><u>helpful</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>you</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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The H⁺ concentration in an aqueous solution at 25 °C is 9.1 × 10⁻⁴. What is [OH⁻]?
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Steps:-

  • First we calculate pH then pOH then [OH-]

\\ \tt\rightarrowtail pH=-log[H^+]

\\ \tt\rightarrowtail pH=-log[9.1\times 10^{-4}]

\\ \tt\rightarrowtail pH=-log9.1-log10^{-4})

\\ \tt\rightarrowtail pH=0.95+4

\\ \tt\rightarrowtail pH=4.95

Now

\\ \tt\rightarrowtail pH+pOH=14

\\ \tt\rightarrowtail pOH=14-4.95

\\ \tt\rightarrowtail pOH=9.05

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\\ \tt\rightarrowtail -log[OH^-]=9.05

\\ \tt\rightarrowtail log[OH^-]=-9.05

\\ \tt\rightarrowtail OH^-=10^{-9.05}

\\ \tt\rightarrowtail OH^-=8.91\times 10^{-4}

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