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Lisa [10]
3 years ago
7

Ice at 0.0 degrees Celsius is combined with 50.0g of water at 75.0 degrees Celsius. Calculate the grams of ice present initially

if the entire mixture comes to a final temperature of 25.00 degrees Celsius after the ice melts. Specific heat of water is 4.18 J/g degrees Celsius. SHOW WORK!
Chemistry
1 answer:
mr_godi [17]3 years ago
4 0

Answer:

m-Ice = 23.85g

Explanation:

The heat absorbed for the cold water (Heat of the increase of water + Heat of ice melting) is equal to the heat released for the water at 75°C. The equation is:

m-Ice-*C*ΔT + Hf*-m-Ice = m-water*C*ΔT

<em>Where m-Ice- is our incognite</em>

<em>C is specific heat of water = 4.18J/g°C</em>

<em>ΔT is change in temperature = 25.0°C - 0.0°C = 25.0°C</em>

<em>Hf is enthalpy of fusion of water = 333.6J/g</em>

<em>m-water = 50.0g</em>

<em>ΔT is change in temperature = 75.0°C - 25.0°C = 50.0°C</em>

<em />

m-Ice-*4.18J/g°C*25°C + 333.6J/g*-m-Ice = 50.0g*4.18J/g°C*50.0°C

104.5 m-Ice- + 333.6 m-Ice- = 10450

438.1 m-Ice = 10450

<h3>m-Ice = 23.85g</h3>

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1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

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