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Veseljchak [2.6K]
3 years ago
13

You are given 100 g of a compound. The compound is composed of 37% hydrogen and 63% oxygen. How many grams of hydrogen are prese

nt in the 100 g sample?
Select one:
a. 37 g
b. 27 g
c. 100 g
d. 63 g
Chemistry
1 answer:
Reil [10]3 years ago
4 0
B is the best answer
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A chemist dissolves of pure potassium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The t
Leya [2.2K]

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

8 0
2 years ago
How do an ionic bond and a covalent bond differ
storchak [24]
<span>An ionic bond is the transfer of electrons and a covalent bond is a sharing of electrons 

Happy studying! ^_^</span>
7 0
3 years ago
I need you to solve for me plzzzzzz
Sergeu [11.5K]

Answer:

.0556 L

Explanation:

First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.

Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).

Finally, the answer will be .0556 L.

<h3 />
3 0
2 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
Details about outer ear
Verdich [7]
Outer ear is the is external part of the ear, consists of the auricle ( also pinna), and the ear canal. It gathers sound energy and focuses it on the ear drum.  
7 0
3 years ago
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