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
The gas flows from higher concentration/pressure to lower concentration/pressure, which is outside the ball.
Answer:
Updrafts characterize a storm's early development, during which warm air rises to the level where condensation begins and precipitation starts to develop. In a mature storm, updrafts are present alongside downdrafts caused by cooling and by falling precipitation.
Hope it helps
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Answer:
2.57 g of H₂
Solution:
The Balance Chemical Equation is as follow,
N₂ + 3 H₂ → 2 NH₃
According to Balance equation,
34.06 g (2 moles) NH₃ is produced by = 6.04 g (3 moles) of H₂
So,
14.51 g of NH₃ will be produced by = X g of H₂
Solving for X,
X = (14.51 g × 6.04 g) ÷ 34.06 g
X = 2.57 g of H₂
