Answer:
T2 = 29.79°C
Explanation:
Equliibrium signifies that heat loss = heat gained
Heat gained by Ice;
H = ML
Mass, M = Number of moles * Molar mass = 1 * 18 = 18g
l = 6.01 k J m o l = 334 J/g
C = 4.186 J/g
H = 18(334)
H = 6012
Heat lost by water
H = MCΔT
H = 18 * 4.186 * (50 - T2)
H = 3767.4 - 75.348T2
Since H = H, we have;
6012 = 3767.4 - 75.348T2
- 75.348T2 = 3767 - 6012
T2 = 2245 / 75.348
T2 = 29.79°C
covalent, because valence electrons are transferred
ionic, because valence electrons are shared
ionic, because valence electrons are transferred
When 3.0 moles of hydrogen peroxide decompose at 1.0 atm and 23°C, approximately 36.78 liters of oxygen gas are produced according to the ideal gas law.
To find the volume of oxygen gas produced when 3.0 moles of hydrogen peroxide decompose at a pressure of 1.0 atm and a temperature of 23°C, you can use the ideal gas law:
PV = nRT
Where:
- P is the pressure (1.0 atm).
- V is the volume (what we want to find).
- n is the number of moles of gas (1.5 moles of O2 since 1 mole of O2 is produced for every 2 moles of H2O2).
- R is the ideal gas constant (approximately 0.0821 L.atm/mol.K).
- T is the temperature in Kelvin (23°C needs to be converted to Kelvin, which is 296.15 K).
First, calculate the number of moles of O2 produced:
n = 3.0 moles of H2O2 * (1 mole of O2 / 2 moles of H2O2) = 1.5 moles of O2
Now, plug in the values into the ideal gas law and solve for V:
1.0 atm * V = 1.5 moles * 0.0821 L·atm/mol·K * 296.15 K
Now, calculate the volume:
V = (1.5 moles * 0.0821 L·atm/mol·K * 296.15 K) / 1.0 atm
Calculate the volume:
V ≈ 36.78 L
So, the volume of oxygen gas produced when 3.0 moles of hydrogen peroxide decompose at 1.0 atm and 23°C is approximately 36.78 liters.
Learn more on ideal gas law here;
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*URGENT
Atoms are the simpliest form of matter. But an atom can be classified as an element and an element can be presented various substance. An element can be oxygen, hydrogen and so on. The combination of two or more elements is a compound. A compound can be a molecule, same as the atom to element.