A attract a core pair of electrons toward itself
Answer:
70.0°C
Explanation:
We are given;
- Amount of heat generated by propane as 104.6 kJ or 104600 Joules
- Mass of water is 500 g
- Initial temperature as 20.0 ° C
We are required to determine the final temperature of water;
Taking the initial temperature is x°C
We know that the specific heat of water is 4.18 J/g°C
Quantity of heat = Mass × specific heat × change in temperature
In this case;
Change in temp =(x-20)° C
Therefore;
104600 J = 500 g × 4.18 J/g°C × (x-20)
104600 J = 2090x -41800
146400 = 2090 x
x = 70.0479
=70.0 °C
Thus, the final temperature of water is 70.0°C
The formula equation for the reaction between hydroiodic acid and beryllium hydroxide is as below
Be(OH)2 + 2 HI = BeI2 + 2H2O
Formula of hydroiodic is = HI
formula of beryllium hydroxide = Be(OH)2
1 mole of Be(OH)2 react with 2 mole of HI to form 1 mole of BeI2 and 2 mole of H2O
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