To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.
We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'

In this, we can see that the unit 'grams' will cancel out to leave us with moles.
In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.
Answer:
(E) changing temperature
Explanation:
Consider the following reversible balanced reaction:
aA+bB⇋cC+dD
If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:
Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)
where:
[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1
There are some important things to remember when calculating Kc:
- <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes
- Pure solids and liquids, including solvents, are not considered for equilibrium expression.
- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc
Answer:
no but that is what we were taught
First we determine the
moles CaCl2 present:
525g / (110.9g/mole) =
4.73 moles CaCl2 present
Based on stoichiometry,
there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>
Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>
Answer:
V=0.68L
Explanation:
For this question we can use
V1/T1 = V2/T2
where
V1 (initial volume )= 0.75 L
T1 (initial temperature in Kelvin)= 303.15
V2( final volume)= ?
T2 (final temperature in Kelvin)= 273.15
Now we must rearrange the equation to make V2 the subject
V2= (V1/T1) ×T2
V2=(0.75/303.15) ×273.15
V2=0.67577931717
V2= 0.68L