Answer:
Option C. 296K
Explanation:
T°C = 23°C
To covert the above to Kelvin temperature, use the following equation
T(K) = T°C + 273
T(K) = 23 + 273
T(K) = 296K
Answer:
There will be produced:
2.97 moles HMnO4
4.45 moles Pb(NO3)2
2.97 moles H2O
Explanation:
Step 1: Data given
Manganese(II) oxide = MnO2
lead(IV) oxide = PbO2
nitric acid = HNO3
Moles of HNO3 = 8.90 moles
Step 2: The balanced equation
2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb(NO3)2 + 2H2O
Step 3: Calculate moles of reactants and products
For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb(NO3)2 and 2 moles of water
For 8.90 moles of HNO3, there will react:
8.90 / 3 = 2.97 moles MnO2
8.90 / 2 = 4.45 moles PbO2
There will be produced:
8.90/3 = 2.97 moles HMnO4
8.90/2 = 4.45 moles Pb(NO3)2
8.90 / 3 = 2.97 moles H2O
Answer:
Chemical energy
Explanation:
When the body eats food our bodies convert the stored energy (calories) to chemical energy, allowing us to function.
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
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