Answer:
positive H and negative S
Explanation:
For a reaction to be spontaneous, the absolute best combination is a negative Delta H and a positive Delta S. When they are both positive, the reaction is only spontaneous at higher temperatures. When they are both negative, the reaction is only spontaneous at lower temperatures. and again if a catalyst is added to the reaction, the activation energy is lowered because a lower-energy transition state is formed. The catalyst does not affect the energy of the reactants or products (and thus does not affect ΔG).
So from these discussions
Ea does not affect G value at all (whether +Ea or -Ea).
And for product to be formed the reaction should be spontaneous, where H is negative and S positive else the reaction will yield low product.
Answer: since the sodium ion is Na+, and sulfate is SO4(2-), you'll need 2 sodiums for a sulfate, making sodium sulfate Na2SO4.
Explanation:
Answer:
513.74 g of solution
Explanation:
% Mass grams are defined as the <em>grams that are dissolved in salt</em> (in this case, it would be <em>potassium nitrate</em>) <em>dissolved every 100 g of the solution</em>. Having this information, you can calculate the amount of solution that has dissolved 18.7 g of potassium nitrate, which is what we want to obtain.
The relationship is:
3.64 g of potassium nitrate _____ 100 g solution
18.7 g of potassium nitrate _____ X = 513.74 g of solution
Calculation: 18.7g x 100g / 3.64g = 513.74 g of solution
So, <em>I need 513.74 g of solution to get 18.7g of potassium nitrate by evaporating it</em>.
Formula for hydroselenic acid: H2Se
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
