Answer:
V H2O = 170.270 mL
Explanation:
- QH2O ( heat gained) = Qcoffe ( heat ceded)
⇒ Q = m<em>C</em>ΔT
∴ m: mass (g)
∴ <em>C</em>:<em> </em>specific heat
assuming:
- δ H2O = δ Coffe = 1.00 g/mL
- <em>C</em> H2O = <em>C</em> coffe = 4.186 J/°C.g....from literature
⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)
∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe
⇒ Q = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J
⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)
⇒ (26371.8 J)/(154.882 J/g) = m H2O
⇒ m H2O = 170.270 g
⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL
Answer:
½O 2 + 2e - + H 2O → 2OH.
Explanation:
Redox reactions - Higher
In terms of electrons:
oxidation is loss of electrons
reduction is gain of electrons
Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:
iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-
oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-
iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-
It would be gold ...................
I would say copper, silver, and tin, since an alloy is a mixture of metals and metalloids.
10. is C or gas and 11. is B liquid.
