Answer:
Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.
Explanation:
Hello,
In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:


The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.
For each case, the reaction quotient is:

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.
Best regards.
I can't really answer his if there is no given. However, there are general rules to follow when naming ionic compounds. For example, if the given compound is written as: MgCl₂.
There are two elements that we could find: Mg and Cl. The rule is to write the positive ion first, followed by the negative ion. The individual charges of each element is the subscript of the other. Since Cl has a subscript of 2, then Mg has a charge of +2. Since Mg has a subscript of 1, then Cl has a charge of -2.
Finally, the rule of naming the ionic compound is to name the ions individually in order, but you have to take the suffix -ide (if the charge is -1). Thus, the name of the compound is Magnesium Chloride,
Answer:
Both b and d can be correct
Explanation:
Generally, diffusion does not require energy (<em>making option a wrong</em>) because it is the movement of particles from a region of high concentration to a region of low concentration hence diffusion moves particles in the direction of a concentration gradient. An example of this is the passive transport (for instance, uptake of glucose by a liver cell).
However, in some cases, when diffusion is against the concentration gradient (i.e when particles move from a region of low concentration to a region of high concentration), diffusion will require energy in a case like this (<em>making option c wrong</em>). An example of this is active transport (transport of protein called sodium-potassium pump which involves pumping of potassium into the cell and sodium out of the cell).
The explanation above shows that diffusion can require energy to move particles (in or out) of the cell through the cell membrane.
Answer:- 27.7 grams of
are produced.
Solution:- The balanced equation is:

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.
Molar mass of
= 207.2+2(126.90) = 461 gram per mol
let's do the calculations for the grams of the product for the given grams of each of the reactant:

= 

= 
From above calculations, NaI gives least amount of
, so the answer is, 27.7 g of
are produced.