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A student collected 48000cm3 of diatomic gas Z at RTP. The RAM of gas Z (MvA=STP) is 24 dm3, RAM=16, L = 6.023x1023. (i) Find the moles of gas Z collected. (ii) Find the mass of gas Z collected. Find the total number of atoms collected.

Chemistry

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3 parts
i

2 mol2 \text{ mol}

ii

64 g64 \text{ g}

iii

2.4092×1024 atomi2.4092 \times 10^{24} \text{ atomi}

Step-by-step solution

6 steps

Haya, twende tukasuluhishe swali hili la kemia.

Taarifa tulizopewa:

  • Kiasi cha gesi ya diatomiki Z₂ iliyokusanywa = 48000 cm348000 \text{ cm}^3 kwenye RTP.
  • Kiasi cha Molar kwenye RTP (MVA=Stp) = 24 dm324 \text{ dm}^3.
  • Masi ya Atomiki Husika (RAM) ya gesi Z = 16.
  • Namba ya Avogadro (L) = 6.023×10236.023 \times 10^{23}.

Part i· 2 steps

Tafuta moli za gesi Z iliyokusanywa.

  1. Badilisha kiasi cha gesi kutoka cm3\text{cm}^3 kwenda dm3\text{dm}^3.

    Tunajua 1 dm3=1000 cm31 \text{ dm}^3 = 1000 \text{ cm}^3.

    Kiasi cha gesi Z2=48000 cm3=480001000 dm3=48 dm3\text{Kiasi cha gesi Z}_2 = 48000 \text{ cm}^3 = \frac{48000}{1000} \text{ dm}^3 = 48 \text{ dm}^3

  2. Tumia fomula ya kupata moli.

    Moli=Kiasi cha gesiKiasi cha Molar kwenye RTP\text{Moli} = \frac{\text{Kiasi cha gesi}}{\text{Kiasi cha Molar kwenye RTP}} Moli za Z2=48 dm324 dm3 mol−1\text{Moli za Z}_2 = \frac{48 \text{ dm}^3}{24 \text{ dm}^3 \text{ mol}^{-1}} Moli za Z2=2 mol\text{Moli za Z}_2 = \boxed{2 \text{ mol}}

Part ii

64 g64 \text{ g}

Tafuta masi ya gesi Z iliyokusanywa.

  1. Tafuta Masi ya Molar ya gesi Z₂.

    Kwa kuwa gesi ni diatomiki (Z₂), na RAM ya Z ni 16:

    Masi ya Molar ya Z2=2×RAM ya Z\text{Masi ya Molar ya Z}_2 = 2 \times \text{RAM ya Z} Masi ya Molar ya Z2=2×16 g mol−1=32 g mol−1\text{Masi ya Molar ya Z}_2 = 2 \times 16 \text{ g mol}^{-1} = 32 \text{ g mol}^{-1}

  2. Tumia fomula ya kupata masi.

    Masi=Moli×Masi ya Molar\text{Masi} = \text{Moli} \times \text{Masi ya Molar} Masi ya Z2=2 mol×32 g mol−1\text{Masi ya Z}_2 = 2 \text{ mol} \times 32 \text{ g mol}^{-1} Masi ya Z2=64 g\text{Masi ya Z}_2 = \boxed{64 \text{ g}}

Part iii

2.4092×1024 atomi2.4092 \times 10^{24} \text{ atomi}

Tafuta jumla ya idadi ya atomi zilizokusanywa.

  1. Tafuta idadi ya molekuli za gesi Z₂.

    Idadi ya molekuli=Moli×Namba ya Avogadro\text{Idadi ya molekuli} = \text{Moli} \times \text{Namba ya Avogadro} Idadi ya molekuli za Z2=2 mol×6.023×1023 molekuli mol−1\text{Idadi ya molekuli za Z}_2 = 2 \text{ mol} \times 6.023 \times 10^{23} \text{ molekuli mol}^{-1} Idadi ya molekuli za Z2=12.046×1023 molekuli\text{Idadi ya molekuli za Z}_2 = 12.046 \times 10^{23} \text{ molekuli}

  2. Tafuta jumla ya idadi ya atomi.

    Kwa kuwa kila molekuli ya Z₂ ina atomi 2 za Z:

    Jumla ya idadi ya atomi=Idadi ya molekuli za Z2×2\text{Jumla ya idadi ya atomi} = \text{Idadi ya molekuli za Z}_2 \times 2 Jumla ya idadi ya atomi=(12.046×1023)×2\text{Jumla ya idadi ya atomi} = (12.046 \times 10^{23}) \times 2 Jumla ya idadi ya atomi=24.092×1023 atomi\text{Jumla ya idadi ya atomi} = 24.092 \times 10^{23} \text{ atomi}

    Au kwa noti ya kisayansi:

    Jumla ya idadi ya atomi=2.4092×1024 atomi\text{Jumla ya idadi ya atomi} = \boxed{2.4092 \times 10^{24} \text{ atomi}}
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