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hichkok12 [17]
3 years ago
10

The electrolysis of water forms H2 and O2.

Chemistry
2 answers:
Monica [59]3 years ago
5 0

Answer:

67.5% ≅ 67.6%

Explanation:

Given data:

Mass of water = 17.0 g

Mass of oxygen produced  (actual yield)= 10.2 g

Percent yield of oxygen = ?

Solution:

Chemical equation:

2H₂O   →   2H₂ + O₂

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 17.0 g/ 18.016 g/mol

Number of moles = 0.944 mol

Now we will compare the moles of oxygen with water to know the theoretical yield of oxygen.

                        H₂O       :        O₂

                          2          :         1

                        0.944      :      1/2×0.944 = 0.472 mol

Mass of oxygen:

Mass = number of moles× molar mass

Mass = 0.472 mol × 32 g/mol

Mass = 15.104 g

Percent yield:

Percent yield = [Actual yield / theoretical yield] × 100

Percent yield = [ 10.2 g/ 15.104 g] × 100

Percent yield = 0.675 × 100

Percent yield = 67.5%

LenaWriter [7]3 years ago
4 0

Answer:

67.6

Explanation:

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Answer:

hydroxide ion

Explanation:

because base produce hydroxide ion when dissolved in water

6 0
3 years ago
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the de
Snezhnost [94]

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P =   ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g   ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

P_{(a)= 790kg/m^3*9.8m/s^2*10           ----------- equation (2)

if we equate equation 1 and 2 together; we have

P_{(g)= P_{(a)

1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm

Making h_g the subject of the formula, we have :

h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}

h_g = 6.269 cm

The difference in the height denoted  by ΔH can therefore be calculated as:

ΔH = H_a-H_g

ΔH = 10cm - 6.269cm

ΔH = 3.731 cm

ΔH ≅ 3.73 cm           (to two significant figures)

5 0
4 years ago
What did we discover with this experiment​
riadik2000 [5.3K]
B is the correct answer
5 0
4 years ago
A rectangle solid of unknown density is 5 m long two meters high and 4 meters wide. The mass of this solid is 300 grams. Find th
aksik [14]
D= M/V so, D=300/40 wich is 7.5 so D= 7.5g/m3
6 0
3 years ago
How many moles of methane are in 7.31*10^25 molecules?
GrogVix [38]
<h3>Answer:</h3>

121 mol CH₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Organic</u>

  • Writing chemical compounds
  • Writing organic structures
  • Prefixes
  • Alkanes, Alkenes, Alkynes

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

7.31 × 10²⁵ molecules CH₄

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u />\displaystyle 7.31 \cdot 10^{25} \ molecules \ CH_4(\frac{1 \ mol \ CH_4}{6.022 \cdot 10^{23} \ molecules \ CH_4} ) = 121.388 \ mol \ CH_4<u />

<u />

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

121.388 mol CH₄ ≈ 121 mol CH₄

7 0
3 years ago
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