Answer:
false thought ia ion of neon = clarity active
Explanation:
Answer:
0.036 M
Explanation:
To do this, let's mark the dye as D and bleach as B.
We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.
The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.
To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:
moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles
moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles
Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:
moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles
These are the moles presents of B after the reaction has been made. The concentration of the same will be:
[B] = 3.597x10⁻⁴ / 0.010
[B] = 0.0357
With 2 SF it would be:
[B] = 3.6x10⁻² M
Answer: 53.3
Explanation:
V2=(T2 x P1 x V1)/(T1 x P2)
(320x50x80)/(300x80)
53.3
Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
