The organism would no longer grow.
Answer:
This question is incomplete.
Explanation:
This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;
Molarity = number of moles (n) / volume (in liter or dm³)
To calculate the number of moles of NaC₂H₃O₂, we say
number of moles (n) =
given or measured mass of NaC₂H₃O₂ ÷ molar mass of NaC₂H₃O₂
The volume of the solvent must be in liter (same as dm³). Thus, to convert mL to liter, we divide by 1000
The unit for Molarity is M (Molar concentration), mol/L or mol/dm³
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
Answer:
The expression of an equilibrium constant will given as:
![K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
K is the constant of a certain reaction when it is in equilibrium

The expression of an equilibrium constant will given as:
![K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)