4.2 g of hydrogen is required to release 1.2 x 10^3 kJ of heat.
Given from the question that 1 mole of hydrogen gas contains 2 g of hydrogen gas.
We also know that heat of combustion per mole of hydrogen gas –285.8 kJ.
From the question, we are told that heat released by the reaction is -1.2 x 10^3 kJ of heat.
Hence;
If 2g of hydrogen releases –285.8 kJ of heat
x g of hydrogen releases -1.2 x 10^3 kJ of heat
x g = 2g x -1.2 x 10^3 kJ / –285.8 kJ
xg = 4.2 g of hydrogen
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Answer:
Food cooking and eating food
Explanation:
You can smell the food, you need to use your eyes to see if its cooked, raw, etc. You need to touch the pan to move it or see if its hot. You can hear the food sizzling or bubbling. And you need to taste the food to see if its good or not
In order to find the two statements, we must first define what the enthalpy of formation and the enthalpy of reaction mean.
Enthalpy of formation:
The change in enthalpy when one mole of substance is formed from its constituent elemetns at standard state.
Enthalpy of reaction:
The change in enthalpy when a reaction occurs and the reactants and products are in their standard states.
Now, we check the statements. The true ones are:
The Hrxn for C(s) + O₂(g) → CO₂(g) is the same as Hf for CO₂
This is true because the formation of carbon dioxide requires carbon and oxygen in their standard states.
The Hf for Br₂<span>(l) is 0 kJ/mol by definition.
Because the bromine is present in its standard state, the enthalpy of formation is 0.
</span><span>The Hrxn for the reaction 1.5H</span>₂<span>(g) + 0.5N</span>₂<span>(g) </span>→ <span>NH</span>₃<span>(g) is the same as the Hf for NH</span>₃<span>(g)
The reactants and products are present in their standard state, and the reaction is the same as the one occurring during the formation of ammonia.
</span>
Orbital notation is a way of writing an electron configuration to provide more specific information about the electrons in an atom of an element.
Orbital notation can be used to determine the quantum numbers of an electron.
Answer:
The cost of electricity for 100 W power bulb = $ 32.85
Cost of electricity for 0.025 W fluorescent bulb = $ 8.2125
Explanation:
Cost of electricity = $ 0.18 per KW-H
Time = 5 hour per day
Bulb power = 100 W = 0.1 KW
Fluorescent bulb power = 25 W = 0.025 KW
(a) Cost of electricity for 100 W power bulb
0.1 × 5 × 365 × 0.18 = $ 32.85
(b) Cost of electricity for 0.025 W fluorescent bulb
0.025 × 5 × 365 × 0.18 = $ 8.2125
Therefore the cost of electricity for 100 W power bulb = $ 32.85
Cost of electricity for 0.025 W fluorescent bulb = $ 8.2125
