<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M
<u>Explanation:</u>
Molarity is calculated by using the equation:
Moles of HI = 0.550 moles
Volume of container = 2.00 L
For the given chemical equation:
<u>Initial:</u> 0.275
<u>At eqllm:</u> 0.275-2x x x
The expression of for above equation follows:
We are given:
Putting values in above expression, we get:
Neglecting the negative value of 'x' because concentration cannot be negative
So, equilibrium concentration of hydrogen gas = x = 0.0275 M
Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M
by classifying into categories that get smaller and smaller these are the classification steps : life , domain , kingdom , phylum , class, order , family , genus , and species every time you narrow it down the option gets smaller and smaller till there's none left except the species it self .
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
Organic chemistry is all about CARBON!!!!!!