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c) Name two major impurities in bauxite.

Chemistry

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Hapa kuna majibu ya maswali yako:

c) Taja uchafu mkuu miwili katika bauxite.
Uchafu mkuu miwili katika bauxite ni:

  • Oksidi za chuma (k.m., Fe2O3\text{Fe}_2\text{O}_3)
  • Silika (SiO2\text{SiO}_2)

16. Wakati wa elektrolisisi ya nitrati ya fedha yenye maji, mkondo wa 5.0A ulipitishwa kwa saa 3.

a) Andika mlinganyo wa mmenyuko uliotokea kwenye kathodi.
Kwenye kathodi, ioni za fedha hupunguzwa na kuunda fedha imara.

Ag+(aq)+e−→Ag(s)\text{Ag}^+(\text{aq}) + \text{e}^- \to \text{Ag}(\text{s})

b) Kokotoa masi ya fedha iliyowekwa kwenye kathodi. (Ag = 108; 1F=96500C).

  1. Kokotoa jumla ya chaji (QQ) iliyopitishwa.

    Muda (tt) = 3 hours=3×60×60 s=10800 s3 \text{ hours} = 3 \times 60 \times 60 \text{ s} = 10800 \text{ s}
    Mkondo (II) = 5.0 A5.0 \text{ A}

    Q=I×t=5.0 A×10800 s=54000 CQ = I \times t = 5.0 \text{ A} \times 10800 \text{ s} = 54000 \text{ C}
  2. Kokotoa idadi ya moli za elektroni.

    Moli za elektroni=QF=54000 C96500 C/mol≈0.559585 mol\text{Moli za elektroni} = \frac{Q}{F} = \frac{54000 \text{ C}}{96500 \text{ C/mol}} \approx 0.559585 \text{ mol}
  3. Kokotoa masi ya fedha iliyowekwa.

    Kutoka mlinganyo wa kathodi, 1 mol ya Ag\text{Ag} huwekwa na 1 mol ya elektroni.
    Hivyo, moli za Ag\text{Ag} = moli za elektroni ≈0.559585 mol\approx 0.559585 \text{ mol}.
    Masi ya molar ya Ag\text{Ag} = 108 g/mol108 \text{ g/mol}.

    Masi ya Ag=Moli za Ag×Masi ya molar ya Ag\text{Masi ya Ag} = \text{Moli za Ag} \times \text{Masi ya molar ya Ag} Masi ya Ag=0.559585 mol×108 g/mol≈60.435 g\text{Masi ya Ag} = 0.559585 \text{ mol} \times 108 \text{ g/mol} \approx 60.435 \text{ g}

    Masi ya fedha iliyowekwa ni 60.44 g\boxed{\text{60.44 g}}.

    17. Kiwanja chenye fomula ya jumla M(OH)₃ humenyuka kama inavyoonyeshwa na mlinganyo hapa chini.
    M(OH)3(s)+OH−(aq)→M(OH)4−(aq)\text{M}(\text{OH})_3(\text{s}) + \text{OH}^-(\text{aq}) \to \text{M}(\text{OH})_4^-(\text{aq})
    M(OH)3(s)+3H+(aq)→M3+(aq)+3H2O(l)\text{M}(\text{OH})_3(\text{s}) + 3\text{H}^+(\text{aq}) \to \text{M}^{3+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

    a) Jina gani hupewa viwanja vinavyoonyesha tabia kama M(OH)₃ katika miitikio miwili?
    Viwanja vinavyoweza kumenyuka kama asidi na besi huitwa amphoteric.

    b) Taja elementi mbili ambazo hidroksidi zao huonyesha tabia kama ile ya M.

    • Aluminium (Al)
    • Zinki (Zn)

    18. a) Eleza sheria ya Graham ya usambaaji.
    Sheria ya Graham ya usambaaji inasema kwamba kiwango cha usambaaji au utoaji wa gesi ni kinyume sawia na mzizi wa mraba wa masi yake ya molar.

    b) Masi za molar za gesi W na X ni 16.0 na 44.0 mtawalia. Ikiwa kiwango cha usambaaji wa W kupitia nyenzo yenye vinyweleo ni 12cm³s⁻¹, kokotoa kiwango cha usambaaji wa X kupitia nyenzo hiyo hiyo.

  4. Andika fomula ya Sheria ya Graham.

    RWRX=MXMW\frac{R_W}{R_X} = \sqrt{\frac{M_X}{M_W}}

    Ambapo RR ni kiwango cha usambaaji na MM ni masi ya molar.

  5. Badilisha maadili yaliyotolewa kwenye fomula.

    RW=12 cm3s−1R_W = 12 \text{ cm}^3\text{s}^{-1}
    MW=16.0M_W = 16.0
    MX=44.0M_X = 44.0

    12RX=44.016.0\frac{12}{R_X} = \sqrt{\frac{44.0}{16.0}}
  6. Rahisisha na utatue kwa RXR_X.

    12RX=2.75\frac{12}{R_X} = \sqrt{2.75} 12RX≈1.6583\frac{12}{R_X} \approx 1.6583 RX=121.6583≈7.236 cm3s−1R_X = \frac{12}{1.6583} \approx 7.236 \text{ cm}^3\text{s}^{-1}

    Kiwango cha usambaaji wa X ni 7.24 cm3s−1\boxed{\text{7.24 cm}^3\text{s}^{-1}}.

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