Answer:
products and reaction
Explanation:
Products
1. FeSO4 and Cu
2. ZnSO4 and Fe
3. CaCl2 and H2
Reactions
1. Fe + CuSO4 → FeSO4 + Cu
2. FeSO4 + Zn → ZnSO4 + Fe
3. 2HCl + Ca → CaCl2 + H2
Answer:
A & B: 8
C: O
D: 26
E: 30
F: Fe
Explanation:
the first one is Oxygen, the other is Iron.
Answer:
An F1 offspring could produce four types of gametes, RY, Ry, rY, and ry. The F2 generation supports the independent-assortment model and refutes the linkage model.
Explanation:
Answer:
The answer is "Option B".
Explanation:
![\to CH_3COOH + NaOH \longleftrightarrow CH_3COONa + H_2O\\\\\to CH_3COONa + NaOH\longleftrightarrow CH3COONa\\\\\therefore \ mol\ NaOH = (5 \ E-3\ L)\times(0.10 \ \frac{mol}{L}) = 5 \ E-4\ mol\\\\](https://tex.z-dn.net/?f=%5Cto%20CH_3COOH%20%2B%20NaOH%20%5Clongleftrightarrow%20%20CH_3COONa%20%2B%20H_2O%5C%5C%5C%5C%5Cto%20CH_3COONa%20%2B%20NaOH%5Clongleftrightarrow%20CH3COONa%5C%5C%5C%5C%5Ctherefore%20%5C%20mol%5C%20%20NaOH%20%3D%20%285%20%5C%20E-3%5C%20%20L%29%5Ctimes%280.10%20%5C%20%5Cfrac%7Bmol%7D%7BL%7D%29%20%3D%205%20%5C%20E-4%5C%20mol%5C%5C%5C%5C)
![\to mol\ CH_3COOH = (0.05 \ L)\times(0.20 \frac{mol}{L}) = 0.01 \ mol\\\\\to C \ CH_3COOH = \frac{(0.01 \ mol - 5 \ E-4\ mol) }{(0.105 \ L)}\\\\\to C \ CH_3COOH = 0.0905 \ M\\\\\therefore \ mol \ CH_3COONa = (0.05\ L )\times (0.20 \ \frac{mol}{L}) = 0.01 \ mol\\\\](https://tex.z-dn.net/?f=%5Cto%20mol%5C%20CH_3COOH%20%3D%20%280.05%20%5C%20L%29%5Ctimes%280.20%20%5Cfrac%7Bmol%7D%7BL%7D%29%20%3D%200.01%20%5C%20mol%5C%5C%5C%5C%5Cto%20C%20%5C%20CH_3COOH%20%3D%20%5Cfrac%7B%280.01%20%5C%20mol%20-%205%20%5C%20E-4%5C%20mol%29%20%7D%7B%280.105%20%5C%20L%29%7D%5C%5C%5C%5C%5Cto%20C%20%5C%20CH_3COOH%20%3D%200.0905%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20%5C%20mol%20%5C%20CH_3COONa%20%3D%20%280.05%5C%20%20L%20%29%5Ctimes%20%280.20%20%5C%20%5Cfrac%7Bmol%7D%7BL%7D%29%20%3D%200.01%20%5C%20mol%5C%5C%5C%5C)
![\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\](https://tex.z-dn.net/?f=%5Cto%20C%20%5C%20CH_3COONa%20%3D%20%20%5Cfrac%7B%280.01%5C%20%20mol%20%2B%205%20%5C%20E-4%5C%20%20mol%29%7D%7B%280.105%5C%20L%20%29%7D%5C%5C%5C%5C%5Cto%20C%20%5C%20CH_3COONa%20%3D%200.1%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20Ka%20%3D%20%28%5BH_3O%5E%7B%2B%7D%5D%5Ctimes%20%5Cfrac%7B%280.1%20%2B%20%5BH_3O%5E%2B%5D%29%29%7D%7B%280.0905%20-%20%5BH_3O%5E%2B%5D%29%7D%20%3D%201.75%5C%20E-5%5C%5C%5C%5C%5Cto%200.1%5BH_3O%5E%2B%5D%20%2B%20%5BH_3O%5E%2B%5D%5E2%20%3D%20%281.75%20E-5%29%5Ctimes%20%280.0905%20-%20%5BH_3O%5E%2B%5D%29%5C%5C%5C%5C)
![\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7](https://tex.z-dn.net/?f=%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%5C%200.1%5BH_3O%5E%2B%5D%20%3D%201.584%5C%20%20E-6%20-%201.75%5C%20%20E-5%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%2B%200.1000175%5BH_3O%5E%2B%5D%20-%201.584%20%5C%20E-6%20%3D%200%5C%5C%5C%5C%5Cto%20%20%5BH_3O%5E%2B%5D%20%3D%201.5835%5C%20%20E-5%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20pH%20%3D%20-%20%5Clog%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%20pH%20%3D%20-%20%5Clog%20%281.5835%20%5C%20E-5%29%5C%5C%5C%5C%20%5Cto%20pH%20%3D%204.8004%20%3E%204.7)