Answer:
Explanation:
We need to use the formula for heat of vaporization.
Identify the variables.
- The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.
- Q is the energy, in this problem, 50,000 Joules.
- m is the mass, which is unknown.
Substitute the values into the formula.
We want to find the mass. We must isolate the variable, m.
m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.
Divide. Note that the Joules (J) will cancel each other out.
Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.
The mass is about 22 grams, so choice B is correct.
Density = mass/volume = 316/22.5 = 14.045g/mL.
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O
<h3>Further explanation</h3>
Given
MnO4- + Br- = MnO2 + BrO3-
Required
The half-reaction
Solution
In acidic conditions :
1. Add the coefficient
2. Equalization O atoms (add H₂O) on the O-deficient side.
3. Equalization H atoms (add H⁺ ) on the H -deficient side. .
4. Equalization of charge (add electrons (e) )
5. Equalizing the number of electrons and then adding the two half -reactions together
Oxidation : Br- → BrO3-
Reduction : MnO4- → MnO2
Br- + 3H2O → BrO3-
MnO4- → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+
MnO4- + 4H+ → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+ + 6e- x1
MnO4- + 4H+ + 3e- → MnO2 + 2H2O x2
- Equalizing the number of electrons and then adding the two half -reactions together
Br- + 3H2O → BrO3- + 6H+ + 6e-
2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O
Br- + 2MnO4- + 3H2O + 8H+ + 6e- → BrO3- + 2MnO2 + 6H+ + 4H2O + 6e-
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O
Water has a high surface tension. Hope that is all you needed=) good luck
Because the easier a metal donates an electron (Low Ionization Energy), the more readily a non-metal with high electromagnetically will accept it.
Thus they match well, - one readily donates, the other readily accepts. This is how ionic bonding occurs.
Hope this helps.