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State one main disadvantage of the process leading to the formation of the brown layer.

Chemistry

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Answer

b
+2

Step-by-step solution

4 steps

Hapa kuna majibu kwa maswali yaliyosalia:

iii) Hasara kuu moja ya mchakato unaosababisha kuundwa kwa safu ya kahawia (kutu) ni kwamba hudhoofisha miundo ya chuma na kupunguza maisha yake ya huduma, na kusababisha gharama za ukarabati au uingizwaji.

4)

Part a

Mlinganyo wa ioni kwa mmenyuko uliotokea kati ya asidi na magnesiamu ni

Mg(s)+2H+(aq)→Mg2+(aq)+H2(g)\text{Mg(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{H}_2\text{(g)}

Part b+2

Tofauti katika ujazo wa gesi iliyotolewa inaelezewa na nguvu tofauti za asidi R na S.

Asidi R ilitoa gesi nyingi zaidi (80 cm3\text{cm}^3) kuliko Asidi S (30 cm3\text{cm}^3) katika muda sawa. Hii inaonyesha kuwa Asidi R ni asidi kali zaidi kuliko Asidi S. Asidi kali huachilia ioni nyingi zaidi za hidrojeni (H+\text{H}^+) kwenye suluhisho, na kusababisha mmenyuko wa haraka na magnesiamu na hivyo kutoa gesi nyingi zaidi ya hidrojeni katika muda uliopimwa.

5) Ili kubaini chaji ya ioni ya P, tutatumia Sheria za Faraday za elektrolisisi.

  1. Badilisha muda kuwa sekunde.

    t=(32 min×60 s/min)+45 s=1920 s+45 s=1965 st = (32 \text{ min} \times 60 \text{ s/min}) + 45 \text{ s} = 1920 \text{ s} + 45 \text{ s} = 1965 \text{ s}

  2. Kokotoa jumla ya chaji (Q) iliyopita.

    Q=I×t=0.5 A×1965 s=982.5 CQ = I \times t = 0.5 \text{ A} \times 1965 \text{ s} = 982.5 \text{ C}

  3. Kokotoa moli za P zilizowekwa.

    Moli za P=Masi ya PRAM ya P=0.44 g88 g/mol=0.005 mol\text{Moli za P} = \frac{\text{Masi ya P}}{\text{RAM ya P}} = \frac{0.44 \text{ g}}{88 \text{ g/mol}} = 0.005 \text{ mol}

  4. Tumia Sheria ya Faraday kuamua chaji ya ioni (nn).

    Tunajua kuwa Q=n×moli za P×FQ = n \times \text{moli za P} \times F, ambapo nn ni chaji ya ioni.

    982.5 C=n×0.005 mol×96500 C/mol982.5 \text{ C} = n \times 0.005 \text{ mol} \times 96500 \text{ C/mol} n=982.50.005×96500n = \frac{982.5}{0.005 \times 96500} n=982.5482.5n = \frac{982.5}{482.5} n≈2.036n \approx 2.036

    Kwa kuwa chaji ya ioni lazima iwe namba kamili, tunazungusha hadi namba kamili iliyo karibu.
    Chaji ya ioni ya P ni +2\boxed{+2}.

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