AH1 = m * c1 * AT1 calculate this for ice (-25C to 0C) AH2 = AHfus(1 mole)=6.01 kJ = 6010 J AH3 = m *c3 * AT3 calculat this for water (0C to 100C) AH4 = AHvap(1mole)=40.67 kJ = 40670 J AH5= m * c5 * AT5 calculate this for steam (100C to 125C)
Sum ---- AH1+AH2+AH3+AH4+AH5
Data m=18g (1mole water)
c1=specific heat ice= 2.09 J/g K c3=specific heat water= 4.18 J/g K c5=specific heat steam= 1.84 J/g K
AT = (Tend - Tinitial) as this is a difference between temperatures it doesn't matter the units Celsius or Kelvin. Kelvin (K)=Celsius (C)+273.15
AT1 = 0C - (-25C)= 25C= 273.15K - 248.15K= 25K AT3= 100C - 0C = 100C= 100K AT5= 125C - 100C= 25C=25K
Answer:
hydrogen atom when it drops from N 5 to N 2?
so, 275 kJ of energy is released when one mole of electrons "falls" from n = 5 to n = 2. E = hc/λ (this energy corresponds to the energy of one photon; the energy calculated in this problem is for one mole of photons so we will change this after we change the units from kJ to J)
Explanation:
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I believe it’s 2? Because it’s saying read the temperature. Which is basically saying look at what state the water is in. Ex: hot or cold
Explanation:
Oxidation state of Nitrogen in N2O5 is +5
Answer:

Explanation:
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In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

So we solve for C2:

2. Now, since 111 mL of water is added, we compute the final volume, V3:

So, the final concentration of the 139 mL portion is:

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