Save the Elephants (STE) is a non-profit organisation. It was set up in 1993 by lain Douglas-Hamilton, and today it is one of the world's largest organisations to save elephants worldwide. It aims to make sure elephants do not die out and protect the habitats in which elephants are found.

Much of the work of STE focuses on stopping the illegal hunting of elephants especially in Africa and Asia, working together with scientists and experts to conduct research into behaviours of elephants, and raising people's awareness through films, televisions and new media sources. So far, it has conducted 335 projects in 40 countries and helped to protect thousands of elephants worldwide.

Phương trình đã cân bằng là:5 CaC2O4+2 KMnO4+8 H2SO4→5 CaSO4+K2SO4+2 MnSO4+10 CO2+8 H2O5\ \text{CaC}_2\text{O}_4 + 2\ \text{KMnO}_4 + 8\ \text{H}_2\text{SO}_4 \rightarrow 5\ \text{CaSO}_4 + K_2\text{SO}_4 + 2\ \text{MnSO}_4 + 10\ \text{CO}_2 + 8\ \text{H}_2\text{O}

1. Tính số mol KMnO₄ đã dùng:

2. soˆˊ mol KMnO₄=CKMnO₄×VKMnO₄=1,00×10−6 mol\text{Số mol KMnO₄} = C_{\text{KMnO₄}} \times V_{\text{KMnO₄}} = 1,00 \times 10^{-6} \, \text{mol}

3. Tính số mol CaC₂O₄ phản ứng:

   Soˆˊ mol CaC₂O₄=52×Soˆˊ mol KMnO₄=2,50×10−6 mol\text{Số mol CaC₂O₄} = \frac{5}{2} \times \text{Số mol KMnO₄} = 2,50 \times 10^{-6} \, \text{mol}

4. Tính số mol Ca²⁺: Mỗi mol CaC₂O₄ giải phóng 1 mol Ca²⁺, nên số mol Ca²⁺ là:

   Soˆˊ mol Ca²⁺=2,50×10−6 molTính nồng độ Ca²⁺ trong máu: Nồng độ ion Ca²⁺ trong mẫu máu 1 mL là:
5. CCa²⁺=2,50×10−3 mol/LC_{\text{Ca²⁺}} = 2,50 \times 10^{-3} \, \text{mol/L}

6. Chuyển đổi sang mg/100 mL:

   CCa²⁺=100,2 mg/100 mLC_{\text{Ca²⁺}} = 100,2 \, \text{mg/100 mL}

Sử dụng công thức để tính enthalpy của phản ứng:

ΔrH2980=[ΔfH2980(Ca2+)+2×ΔfH2980(Cl−)]−ΔfH2980(CaCl2)

Thay giá trị vào:
\Delta_r H_{298}^0 = [(-542,83) + 2 \times (-167,16)] - (-795,0)Δr​H2980​=[(−542,83)+2×(−167,16)]−(−795,0)
\Delta_r H_{298}^0 = [-542,83 - 334,32] + 795,0Δr​H2980​=[−542,83−334,32]+795,0
\Delta_r H_{298}^0 = -877,15 + 795,0Δr​H2980​=−877,15+795,0
\Delta_r H_{298}^0 = -82,15 \text{ kJ/mol}Δr​H2980​=−82,15 kJ/mol

a,

•  Fe + HNO3 → Fe(NO3)3 + NO + H2O

  • Chất oxi hóa: HNO3 (NO₃⁻).
  • Chất khử: Fe.
  • Quá trình oxi hóa: Fe → Fe³⁺ (mất 3 electron).
  • Quá trình khử: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H2O.

•  

b,