Answer:
![1.41 *10^{3} kJ/mol](https://tex.z-dn.net/?f=1.41%20%2A10%5E%7B3%7D%20%20kJ%2Fmol)
Explanation:
First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)
Then we solve the ecuation for ΔH°reaction
ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)
ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol
ΔH°reaction = -1.41 *10^3 kJ/mol
Answer:
![\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B10%20%5Cpi%7D)
Explanation:
Volume of a cone:
We have
and we want to find
when the height is 2 cm.
We can see in our equation for the volume of a cone that we have three variables: V, r, and h.
Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.
We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.
Plug this value for r into the volume formula:
Differentiate this equation with respect to time t.
Plug known values into the equation and solve for dh/dt.
Divide both sides by 100π to solve for dh/dt.
The height of the cone is increasing at a rate of 1/10π cm per second.
Answer:
Increase pressure 3X => increase Temperature 3X
Explanation:
Gay-Lussac Law => T ∝ P => T =kP => Empirical Relationship => T₁/P₁ = T₂/P₂
=> T₂ = T₁P₂/P₁
Given P₂ = 3P₁ => T₂ = T₁(3P₁)/P₁ = 3T₁
Answer:
17.6 moles of oxygen gas.
Explanation:
STP Variables:
P=1 atm
R=0.082
T=273 K
Use the PV=nRT, then plug in.
n=PV/RT
n= (1 atm)(395 L)/(0.082) (273 K)
Simplify.
n=17.6 moles of O2
Hoped this helped.