Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion + 
Where, negative sign signifies heat loss
Or,
Heat of fusion + 
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,


345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>
1) Excess reagent
1 mol N2 / 3 mol H2
6.0 mol N2 *3 mol H2 / 1 mol N2 = 18 mol H2
18mol H2 > 12 mol H2 => H2 is limiting (you need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.
12.0 mol H2 * 1mol N2/ 3 mol H2 = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2
2) NH3 produced
12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3
Aslso, 4 mol N2 *[2molNH3 / 1 molN2] = 8 mol NH3, the same result.
3) Yield
80% * 8 mol NH3 = 6.4 mol NH3
6.02 × 10^23 hydrogen ions per mol
Answer:
Esto se debe a la rotación de la tierra
Explanation:
Answer: C. Light can cause electrons to be released from the surface of a metal
Explanation: