Answer:
C is the element thats has been oxidized.
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)
This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.
In the oxalic acid, this are the oxidation states:
H: +1
C: +3
O: -2
In the product side, in CO₂ the oxidation states are:
C: +4
O: -2
Carbon from the oxalate has increased the oxidation state, so it has been oxidized.
Answer:
1) The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water. Notice that enthalpy change of neutralization is always measured per mole of water formed. Enthalpy changes of neutralization are always negative - heat is released when an acid and and alkali react. For reactions involving strong acids and alkalis, the values are always very closely similar, with values between -57 and -58 kJ mol-1. That varies slightly depending on the acid-alkali combination (and also on what source you look it up in!).
2)same of n1
Explanation:
oxygen (O)will react with water(H2O) to form hydroxide which is (OH
<h3>Answer: </h3>
The answer is gold.
<h3>Explanation:</h3>
Substance: Specific Heat (C) in joules per gram/degrees centigrade:
Water (ice) 2.05
Iron 0.46
Aluminium 0.90
Gold 0.13
Copper 0.39
Ammonia (liquid) 4.70
Ethanol 2.44
Gasoline 2.22
Water (liquid) 4.18
Water (vapor) 2.08
Air (25 degrees Celsius) 1.01
Oxygen 0.92
Hydrogen 14.30
To know that we clear the equation q=m*C*Δt for the specific heat (C) and solve the equation. Keep in mind that we need to use the heat (520 J) in Joules since the values from the question and the table are in Joules.
q/m∗ΔT = C = 520J/200g∗(45degC−25degC) =
0.13J/gdegC
As you can see, the substance that has this specific heat is GOLD.
