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Name i) the process shown by the equation above

Chemistry

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Haya hapa majibu ya maswali uliyotuma:

18. i)

  1. Pata masi ya solid P kavu.

    Masi ya solid P kavu = (Masi ya chombo cha kuyeyushia + solid P kavu) - (Masi ya chombo cha kuyeyushia tupu)

    =28.4 g−24.2 g= 28.4 \text{ g} - 24.2 \text{ g} =4.2 g= 4.2 \text{ g}
  2. Pata masi ya maji katika mmumunyo uliokolea.

    Masi ya mmumunyo uliokolea = (Masi ya chombo cha kuyeyushia + mmumunyo uliokolea) - (Masi ya chombo cha kuyeyushia tupu)

    =40.4 g−24.2 g= 40.4 \text{ g} - 24.2 \text{ g} =16.2 g= 16.2 \text{ g}

    Masi ya maji = Masi ya mmumunyo uliokolea - Masi ya solid P kavu

    =16.2 g−4.2 g= 16.2 \text{ g} - 4.2 \text{ g} =12.0 g= 12.0 \text{ g}
  3. Kokotoa umumunyifu wa solid P kwa 100 g ya maji.

    Umumunyifu = Masi ya solid PMasi ya maji×100 g\frac{\text{Masi ya solid P}}{\text{Masi ya maji}} \times 100 \text{ g}

    =4.2 g12.0 g×100 g= \frac{4.2 \text{ g}}{12.0 \text{ g}} \times 100 \text{ g} =0.35×100 g= 0.35 \times 100 \text{ g} =35 g= 35 \text{ g}

    Umumunyifu wa solid P ni 35 g kwa 100 g ya maji\boxed{\text{35 g kwa 100 g ya maji}}.

    18. ii)
    Tahadhari moja iliyozingatiwa ni:

    • Pasha moto kwa upole ili kuzuia mchanganyiko kutapakaa.

    19. i)
    Hali mbili zinazoharakisha kutu ni:

    • Kuwepo kwa maji ya chumvi.
    • Joto la juu.

    19. ii)
    Njia mbili za kuzuia kutu ni:

    • Kupaka rangi.
    • Kupaka mafuta/grisi.

    20. i)
    Mchakato ulioonyeshwa na mlinganyo hapo juu unaitwa:

    • Ufaaji (Cracking).
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