Answer:
Sugar, sodium chloride, and hydrophilic proteins
What you can do is organize them by color, what matter they are in room temperature, their molecular structure, or what kind of conductor in electricity and heat it is. I'm not sure what the format is supposed to look like but first just organize them all in different categories.
A large number in front of a compound designates how many units there are of that compound. Parentheses can be used to designate a special structure, where other molecules are attached to the larger, complex molecule.
Answer:
The equation that gives the overall equilibrium in terms of the equilibrium constants K and Ky is K1 = K^6 * Ky
Explanation:
we have the following balanced reaction:
CaC2 + 2H2O = C2H2 + Ca(OH)2
the value of K for this reaction will be equal to:
K = ([C2H2] * [Ca(OH)2])/([CaC2] * [H2O]^2)
if we multiply the reaction by the value of 6, we have:
6CaC2 + 12H2O = 6C2H2 + 6Ca(OH)2
Again, the value of K for this reaction will be equal to:
K,´ = ([C2H2] ^6 * [Ca(OH)2]^6)/([CaC2]^6 * [H2O]^12) = K^6
For the second reaction:
6C2H2 + 3CO2 + 4H2O = 5CH2CHCO2H
The value of K for this reaction:
K2 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^4)
we also have:
K1 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^16)
Thus:
K1 = K^6 * Ky
Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99
