Find the inverse of the matrix

$$\frac{G_2}{G_1} = \frac{G_3}{G_2} \implies \frac{a_A + 12}{a_A + 6} = \frac{a_A + 21}{a_A + 12}$$ $$(a_A + 12)^2 = (a_A + 6)(a_A + 21)$$ $$a_A^2 + 24a_A + 144 = a_A^2 + 21a_A + 6a_A + 126$$ $$a_A^2 + 24a_A + 144 = a_A^2 + 27a_A + 126$$ $$24a_A + 144 = 27a_A + 126$$ $$144 - 126 = 27a_A - 24a_A$$ $$18 = 3a_A$$ $$a_A = 6$$

i) Neno la kwanza la GP.
Neno la kwanza la GP ni $G_1 = a_A + 6 = 6 + 6 = \boxed{\text{12}}$.

$$r_{GP} = \frac{G_2}{G_1} = \frac{a_A + 12}{a_A + 6} = \frac{6 + 12}{6 + 6} = \frac{18}{12} = \frac{3}{2}$$

ii) Jumla ya maneno 9 ya kwanza ya GP kwa tarakimu 4 muhimu.
Fomula ya jumla ya maneno $n$ ya kwanza ya GP ni $S_n = \frac{G_1(r_{GP}^n - 1)}{r_{GP} - 1}$.

$$S_9 = \frac{12\left(\left(\frac{3}{2}\right)^9 - 1\right)}{\frac{3}{2} - 1}$$ $$S_9 = \frac{12\left(\frac{19683}{512} - 1\right)}{\frac{1}{2}}$$ $$S_9 = 24\left(\frac{19683 - 512}{512}\right)$$ $$S_9 = 24\left(\frac{19171}{512}\right)$$ $$S_9 = \frac{450096}{512} = 879.09375$$

Jumla ya maneno 9 ya kwanza ya GP kwa tarakimu 4 muhimu ni $\boxed{\text{879.1}}$.

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Sasa, tutatatua maswali ya kemia kutoka kwenye picha uliyotuma.

Swali la 1:
Gesi ya nitrojeni IV oksidi ($NO_2$) husambaa $100\, \text{cm}^3$ kwa sekunde 45. Tunataka kujua itachukua muda gani kwa kiasi sawa cha gesi ya salfa IV oksidi ($SO_2$) kusambaa.
Uzito wa molekuli wa $NO_2$: $M_{NO_2} = 14 + (2 \times 16) = 14 + 32 = 46\, \text{g/mol}$.
Uzito wa molekuli wa $SO_2$: $M_{SO_2} = 32 + (2 \times 16) = 32 + 32 = 64\, \text{g/mol}$.
Tunatumia Sheria ya Graham: $\frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}}$.
Kwa kuwa kiasi ni sawa, $\frac{V/T_1}{V/T_2} = \frac{T_2}{T_1}$. Hivyo, $\frac{T_{SO_2}}{T_{NO_2}} = \sqrt{\frac{M_{SO_2}}{M_{NO_2}}}$.

$$\frac{T_{SO_2}}{45\, \text{s}} = \sqrt{\frac{64\, \text{g/mol}}{46\, \text{g/mol}}}$$ $$\frac{T_{SO_2}}{45} = \sqrt{1.3913}$$ $$\frac{T_{SO_2}}{45} \approx 1.1795$$

$$T_{SO_2} = 45 \times 1.1795$$ $$T_{SO_2} \approx 53.0775\, \text{s}$$

Itachukua takriban $\boxed{\text{53.1\, s}}$ kwa $SO_2$ kusambaa.

Swali la 2:
Gesi V husambaa kwa kiwango cha $10\, \text{cm}^3/\text{s}$ na gesi W husambaa kwa kiwango cha $10.69\, \text{cm}^3/\text{s}$. Uzito wa molekuli wa W ni 28. Tafuta uzito wa molekuli wa V.

$$\frac{Rate_V}{Rate_W} = \sqrt{\frac{M_W}{M_V}}$$ $$\frac{10\, \text{cm}^3/\text{s}}{10.69\, \text{cm}^3/\text{s}} = \sqrt{\frac{28}{M_V}}$$ $$0.93545 \approx \sqrt{\frac{28}{M_V}}$$

$$(0.93545)^2 = \frac{28}{M_V}$$ $$0.87506 \approx \frac{28}{M_V}$$

$$M_V = \frac{28}{0.87506}$$ $$M_V \approx 31.997$$

Uzito wa molekuli wa gesi V ni takriban $\boxed{\text{32}}$.

Swali la 3:
$50\, \text{cm}^3$ ya gesi ya salfa IV oksidi ($SO_2$) husambaa kwa sekunde 20. Ni kiasi gani cha gesi ya nitrojeni IV oksidi ($NO_2$) kitasambaa kwa sekunde 10?
Uzito wa molekuli wa $SO_2$: $M_{SO_2} = 64\, \text{g/mol}$.
Uzito wa molekuli wa $NO_2$: $M_{NO_2} = 46\, \text{g/mol}$.

$$\frac{Rate_{SO_2}}{Rate_{NO_2}} = \sqrt{\frac{M_{NO_2}}{M_{SO_2}}}$$ $$\frac{2.5}{Rate_{NO_2}} = \sqrt{\frac{46}{64}}$$ $$\frac{2.5}{Rate_{NO_2}} = \sqrt{0.71875}$$ $$\frac{2.5}{Rate_{NO_2}} \approx 0.84779$$

$$\text{Kiasi}_{NO_2} = Rate_{NO_2} \times Time$$ $$\text{Kiasi}_{NO_2} = 2.9499\, \text{cm}^3/\text{s} \times 10\, \text{s}$$ $$\text{Kiasi}_{NO_2} \approx 29.499\, \text{cm}^3$$

Kiasi cha $NO_2$ kitakachosambaa ni takriban $\boxed{\text{29.5\, cm}^3}$.

Swali la 4:
Kiasi fulani cha gesi ya $SO_2$ huchukua sekunde 80 kusambaa. Kiasi sawa cha gesi X huchukua sekunde 40 kusambaa. Tafuta R.M.M. ya gesi X.
Uzito wa molekuli wa $SO_2$: $M_{SO_2} = 64\, \text{g/mol}$.
Kwa kuwa kiasi ni sawa, $\frac{T_X}{T_{SO_2}} = \sqrt{\frac{M_X}{M_{SO_2}}}$.

$$\frac{40\, \text{s}}{80\, \text{s}} = \sqrt{\frac{M_X}{64\, \text{g/mol}}}$$ $$0.5 = \sqrt{\frac{M_X}{64}}$$

$$(0.5)^2 = \frac{M_X}{64}$$ $$0.25 = \frac{M_X}{64}$$

$$M_X = 0.25 \times 64$$ $$M_X = 16$$

R.M.M. ya gesi X ni $\boxed{\text{16}}$.

Swali la 5:
$100\, \text{cm}^3$ ya gesi ya nitrojeni ($N_2$) husambaa kwa sekunde 40. Itachukua muda gani kwa $100\, \text{cm}^3$ ya gesi ya oksijeni ($O_2$) kusambaa?
Uzito wa molekuli wa $N_2$: $M_{N_2} = 2 \times 14 = 28\, \text{g/mol}$.
Uzito wa molekuli wa $O_2$: $M_{O_2} = 2 \times 16 = 32\, \text{g/mol}$.
Kwa kuwa kiasi ni sawa, $\frac{T_{O_2}}{T_{N_2}} = \sqrt{\frac{M_{O_2}}{M_{N_2}}}$.

$$\frac{T_{O_2}}{40\, \text{s}} = \sqrt{\frac{32\, \text{g/mol}}{28\, \text{g/mol}}}$$ $$\frac{T_{O_2}}{40} = \sqrt{1.142857}$$ $$\frac{T_{O_2}}{40} \approx 1.0690$$

$$T_{O_2} = 40 \times 1.0690$$ $$T_{O_2} \approx 42.76\, \text{s}$$

Itachukua takriban $\boxed{\text{42.8\, s}}$ kwa oksijeni kusambaa.