Answer: B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O (g) (ΔH = -2035 kJ/mol) 3H2O (g) → 3H2O (l) (ΔH = -132 kJ/mol) 3H2O (l) → 3H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)
Explanation: ??
Answer:
homogeneous mixtures: iron,alcohol,zonrox,wine.
heterogeneous mixtures: smoke,batchoy,spaghetti,halo halo,book,clothes.
Answer:
C
Explanation:
This is a correct answer i think soo
Answer:
- <em>The maximum amount of copper allowed in 100 g of water is </em><u><em>0.00013 g</em></u>
Explanation:
To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg) and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):
First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.
Now, set the proportion:
- 1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O
Solve for x:
- x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu
Convert mg to grams:
- 0.13 mg × 1 g / 1,000 mg = 0.00013 g
Answer: 0.00013 g of copper.