Part 1: Fill in the blank Newton’s Second Law: Unbalanced forces cause an object to _____

(LO 3N, 4G, 4O) Aluminum sulfate is also involved in dying fabrics. The gelatinous precipitate formed in the reaction with dilut

brilliants [131]

9.4 × 10⁻³ mg (0.73 mmoles) of Al(OH)₃ is formed

Explanation:

We have the following chemical reaction:

Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(s) + 3 Na₂SO₄(aq)

The precipitate mentioned by the problem is aluminium hydroxide Al(OH)₃.

Now to determine the number of moles of sodium hydroxide NaOH we use the following formula:

molar concentration = number of moles / volume

number of moles = molar concentration × volume

number of moles of NaOH = 0.088 M × 25 mL = 2.2 mmoles

number of moles of Al₂(SO₄)₃ = 5.6 × 10⁻³ moles = 5.6 mmoles (found in the problem text)

We see from the chemical reaction that 1 mole of Al₂(SO₄)₃ requires 6 moles of NaOH so 5.6 mmoles of Al₂(SO₄)₃ would require 6 times more NaOH which is 33.6 mmoles and we have only 2.2 mmoles. The limiting reactant will be NaOH.

Now we devise the following reasoning:

if 6 mmoles of NaOH produces 2 mmoles of Al(OH)₃

then 2.2 mmoles of NaOH produces X mmoles of Al(OH)₃

X = (2.2 × 2) / 6 = 0.73 mmoles of Al(OH)₃

mass of Al(OH)₃ = number of moles / molecular weight

mass of Al(OH)₃ = 0.73 / 78

mass of Al(OH)₃ = 9.4 × 10⁻³ mg

Learn more:

precipitation reaction

brainly.com/question/10400269

7 0

3 years ago

1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta

kompoz [17]

The answers to the two questions are:

1. The percent abundance in the container which has 100 navy, 27 pinto, and 173 black-eyed peas beans is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The weighted average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.

1. The percent abundance by type of bean is given by:

$\% = \frac{n}{n_{t}} \times 100$ (1)

Where:

n: is the number of each type of beans

$n_{t}$ : is the total number of beans

The total number of beans can be calculated by adding the number of all the types of beans:

$n_{t} = n_{n} + n_{p} + n_{b}$ (2)

Where:

$n_{n}$ : is the number of navy beans = 100

$n_{p}$ : is the number of pinto beans = 27

$n_{b}$ : is the number of black-eyed peas beans = 173

Hence, the total number of beans is (eq 2):

$n_{t} = 100 + 27 + 173 = 300$

Now, the percent abundance by type of bean is (eq 1):

Navy beans

$\%_{n} = \frac{100}{300} \times 100 = 33.3 \%$

Pinto beans

$\%_{p} = \frac{27}{300} \times 100 = 9.0 \%$

Black-eyed peas beans

$\%_{b} = \frac{173}{300} \times 100 = 57.7 \%$

Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The average score (S) can be calculated as follows:

$S = e*\%_{e} + l*\%_{l} + h*\%_{h}$ (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

$%_{e}$ $\%_{e}$ : is the percent for exams = 70.0%

$\%_{l}$ : is the percent for lab reports = 20.0%

$\%_{h}$ : is the percent for homework = 10.0%

Then, the average score is:

$S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1$

We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.

Therefore, the weighted average score will be 84.1.

Find more about percents here:

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I hope it helps you!

4 0

3 years ago

The chemical formula for a molecule that contains two chlorine (Cl) atoms is ___________.

Karo-lina-s [1.5K]

CI2 is two chlorine atoms.

6 0

4 years ago

Read 2 more answers

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid. MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l)

Marina CMI [18]

Answer:

Please see the complete formt of the question below

Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.

MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)

According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.

The answer to the above question is

The limiting reactant is the MnO₂

Explanation:

To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂

Molar mass of MnO₂ = 86.9368 g/mol

Molar mass of HCl = 36.46 g/mol

From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂

However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed

7 0

3 years ago

If 30.0 grams of sulfuric acid react with 26.0 grams of aluminum hydroxide in a double replacement reaction, how many grams of w

FromTheMoon [43]

Answer:

a) mass of water produced= 11.16 g of water

b) mass of excess reactant= 9.36 g of aluminum hydroxide

Explanation:

The balanced reaction equation is;

3H2SO4 + 2Al(OH)3 → 6H2O + Al2(SO4)3

The next step is to determine the limiting reactant. The limiting reactant will give the least number of moles of product.

For sulphuric acid;

Molar mass of sulphuric acid =98gmol-1

Number of moles of sulphuric acid= 30g/98gmol-1 = 0.31 moles of sulphuric acid

From the reaction equation;

3 moles of sulphuric acid yields 6 moles of water

0.31 moles of sulphuric acid will yield 0.31 ×6/3 = 0.62 moles of water

For aluminum hydroxide;

Number of moles= 26.0 g/78 g/mol = 0.33 moles

From the reaction equation;

2 moles of aluminum hydroxide yields 6 moles of water

0.33 moles of aluminum hydroxide will yield 0.33 × 6/2 = 0.99 moles of water

Hence sulphuric acid is the limiting reactant.

Thus mass of water produced= 0.62 moles ×18gmol-1 = 11.16 g of water

Since 3 moles of sulphuric acid reacts with 2 moles of aluminum hydroxide

0.31 moles of sulphuric acid reacts with 0.31 ×2/3 = 0.21 moles of aluminum hydroxide

Thus amount of excess reactant = 0.33- 0.21 = 0.12 moles of aluminum hydroxide

Mass of excess aluminum hydroxide = 0.12 × 78 g/mol = 9.36 g of aluminum hydroxide

4 0

3 years ago

Part 1: Fill in the blank Newton’s Second Law: Unbalanced forces cause an object to ______.