Answer:
0.56 g
Explanation:
<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>
Step 1: Given data
Moles of nitrogen gas (n): 0.020 mol
Step 2: Calculate the molar mass (M) of nitrogen gas
Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:
M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol
Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas
We will use the following expression.
m = n × M
m = 0.020 mol × 28.02 g/mol
m = 0.56 g
Oceanic crust would be on top, being less dense and doesn't have as much water in it. Old oceanic crust is usually on the bottom, and filled with water. It is more dense.
Answer:
explain what need to be done here pls? what do you need?
Explanation:
Answer:
The three blanks for this answer, are
1. volumen
2. moles
3. Temperature and pressure.
So, Avogadro's law states that the volume of a gas is directly proportional to the moles of the gas when temperature and pressure stay the same
Explanation:
Imagine you have 10 moles of a gas which is contained in 50 L. How many moles of that gas, you will have if the volumen has been reduced to 10 L. (Of course, don't forget that T° and pressure are the same)
There is an equation like this, initial moles /initial volume = moles at the end/volume at the end, (Avogadro law for gases), so 10/50 =moles at the end/10. When u operate, moles at the end = (10 x 10) / 50.
Moles at the end are 2. Did u get it?. Volumen has been reduced, also the moles.
Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
