If you me ill give you mores. plzz .at 1 atm, how much energy is required to heat 37.0 g of h2o(s) at –10.0 °c to h2o(g) at 155.0 °c? constants can be found here.

116.3 kJ

Step-by-step explanation:

Three heat transfers are involved

q = Heat to warm ice + heat to melt ice + heat to warm water + heat to evaporate water + heat to warm steam

q =      q₁      +     q₂        +     q₃       +     q₄          +     q₅

q = mC₁ΔT₁ + mΔH_fus + mC₃ΔT₃ + mΔH_vap + mC₅ΔT₅

Step 1: Calculate q₁

m = 37.0 g

C₁ = 2.010 J·°C⁻¹g⁻¹

ΔT₁ = T_f – T_i  

ΔT₁ = 0.0 – (-10.0)

ΔT₁ = 10.0 °C  

q₁ = 37.0 × 2.010 × 10.0  

q₁= 743.7 J

q₁= 0.7437 kJ

Step 2. Calculate q₂

ΔH_fus = 334 J/g

q₂ = 37.0 × 334

q₂ = 12 360 J

q₂ = 12.36 kJ

Step 3: Calculate q₃

C₃ = 4.179 J·°C⁻¹g⁻¹

ΔT₃ = T_f – T_i  

ΔT₃ = 100 – 0  

ΔT₃ = 100 °C

q₃ = 37.0 × 4.179 × 100  

q₃ = 15 460 J

q₃ = 15.46 kJ

Step 4. Calculate q₄

ΔH_vap = 2260 J/g

q₄ = 37.0 × 2260

q₄ = 83 620 J

q₄ = 83.62 kJ

Step 5. Calculate q₅

C¬₅ = 2.010 J·°C⁻¹g⁻¹

ΔT₅ = T_f – T_i  

ΔT₅ = 155.0 – 1000  

ΔT₅ = 55.0 °C

q₅ = 37.0 × 2.010 × 55

q₅ = 4090 J

q₅ = 4.090 kJ

Step 6. Calculate q

q = 0.7437 + 12.36 + 15.46 + 83.62 + 4.090

q = 116.3 kJ

The heat required is 116.3 kJ.

false

because it does not have the potential to be entirely used up

elements make up everything, and they combine to make compounds.

the answer is c