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myrzilka [38]
3 years ago
12

Calculate the energy needed to heat 10.5 g ice at -20.0 °C to liquid water at 75.0 °C. The heat of vaporization of water = 2257

J/g, the heat of fusion of water = 334 J/g, the specific heat capacity of water = 4.18 J/g·°C, and the specific heat capacity of ice = 2.06 J/g·°C.
Chemistry
1 answer:
lawyer [7]3 years ago
3 0

Answer:

= 7234.5Joules

Explanation:

Q = (mcΔT)ice + (mΔH)melting +(mcΔT)water

= (10.5g)(2.06J/g°C)[0°C-(-20°C)+(10.5g)(2257J/g)+(10.5g)(4.184J/g°C(75°C - 0°C)

= [(10.5)(2.06)(20) + (10.5)(2257) + (10.5)(4.184)(75)]J

= 432.6J + 3507J + 3294.9J

= 7234.5Joules

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A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de
zhannawk [14.2K]

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

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p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

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\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

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Answer:

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Let's introduce a strategy needed to solve any similar problem like this:

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