Chemistry College

Calculate the volume of a box which is 125 cm long, 37 cm wide, and 68 cm high. Report your answer with correct significant figures in cubic centimeters. please use significant figures!

Calculate the volume of a box which is 423 cm long, 12 cm wide, and 25 cm high. have significant figures in cubic centimeters.

Suppose you are measuring the mass of a solid sample on a balance using a weigh boat. You record the data in a table. Mass of weigh boat 1.326 g Mass of weigh boat and sample 7.635 g What is the mass of the solid sample?

A graduated cylinder contains 10.00 mL water. A 14.74 g piece of aluminum is added to the water, and the volume rises to 15.46 mL. What is the density of the aluminum?

A 24.5 g sample of solution has a density of 0.768 g/mL. What is the sample volume (in mL)?

Answers

Answer 1

Answer:

1. Volume of box is 314500 cm³.

2. Volume of box 126900 cm³.

3. Mass of sample is 6.309 g.

4. Density of aluminum is 2.7 g/mL.

5. Volume of sample is 31.9 mL.

Explanation:

1. Determination of the volume of the box

Length (L) = 125 cm

Width (W) = 37 cm

Height (H) = 68 cm

Volume (V) =?

V = L × W × H

V = 125 × 37 × 68

V = 314500 cm³

2. Determination of the volume of the box

Length (L) = 423 cm

Width (W) = 12 cm

Height (H) = 25 cm

Volume (V) =?

V = L × W × H

V = 423 × 12 × 25

V = 126900 cm³

3. Determination of the mass of the sample.

Mass of weigh boat = 1.326 g

Mass of weigh boat + Sample = 7.635 g Mass of the sample =?

Mass of Sample = (Mass of weigh boat + Sample) – (Mass of weigh boat)

Mass of Sample = 7.635 – 1.326

Mass of Sample = 6.309 g

4. Determination of the density of aluminum.

Volume of water = 10 mL

Mass of aluminum = 14.74 g

Volume of water + aluminum = 15.46 mL

Density of aluminum =?

Next, we shall determine the volume of aluminum. This can be obtained as follow:

Volume of water = 10 mL

Volume of water + aluminum = 15.46 mL

Volume of aluminum =?

Volume of aluminum = (Volume of water + aluminum) – (Volume of water)

Volume of aluminum = 15.46 – 10

Volume of aluminum = 5.46 mL

Finally, we shall determine the density of aluminum. This can be obtained as follow:

Mass of aluminum = 14.74 g

Volume of aluminum = 5.46 mL

Density of aluminum =?

Density = mass / volume

Density of aluminum = 14.74 / 5.46

Density of aluminum = 2.7 g/mL

5. Determination of the volume of sample.

Mass of sample = 24.5 g

Density of sample = 0.768 g/mL.

Volume of sample =?

Density = mass / volume

0.768 = 24.5 / volume of sample

Cross multiply

0.768 × Volume of sample = 24.5

Divide both side by 0.768

Volume of sample = 24.5 / 0.768

Volume of sample = 31.9 mL.

Related Questions

Do the substances that are reactants in a chemical reaction remain the same after the reaction is complete?

Answers

The first answer is nine you are currently correct

atomic radius(size)______ from left to right across a period on the periodic table

Answers

Answer:

Decreases

Explanation:

The mass of an unidentified rock is 15.5 grams. Students determine the volume of the rock by placing the rock in a cylinder with water. What was the density of the rock that the students should have calculated?

Hint: The graduated cylinder measures volume. Read the difference in water level.

Answers

The concept of water by displacement. D=m/v therefore m=15.5 g and volume=5 mL (final - initial) then calculate the density

Which factor can help balance Earth’s temperature?

Answers

Answer:

Clouds

Explanation:

Because I just got it right

Determine the mass of oxygen that is needed to react with 33 grams of phosphorus.
4P + 5 O2 → 2 P205
A
27.3 grams O2
B
1.33 grams O2
С
213 grams 02
D
42.6 grams 02

Answers

Answer:

D =  42.6 g

Explanation:

Given data:

Mass of oxygen needed = ?

Mass of phosphorus react = 33 g

Solution:

Chemical equation:

4P + 5O₂     →      2P₂O₅

Number of moles of phosphorus:

Number of moles = mass/molar mass

Number of moles = 33 g/ 30.97 g/mol

Number of moles = 1.065 mol

now we will compare the moles of P and oxygen.

                P         :       O₂

                4         :        5

          1.065        :        5/4×1.065 = 1.33mol

Mass of oxygen needed:

Mass = number of moles × molar mass

Mass = 1.33 mol × 32 g/mol

Mass = 42.6 g

What type of charge do ionic compounds have

Answers

Answer:neutral

Explanation:ionic compounds is between a cation and anion so they cancel eachother out

The blood cells that directly attack germs are: *​

Answers

Answer:

lymphocytes

Explanation:

it consists of the b and the t cells

The white blood cells are the ones that attack germs.

A 558 mg of a mixture of fluorene and benzoic acid was weighed out and subjected to an extraction and recrystallization. After this purification was completed the product crystals were dried and analyzed. The purification procedure produced 185 mg of fluorene and 144 mg of benzoic acid. Calculate the percent composition of this mixture.

Answers

Answer:

See explanation

Explanation:

Mass of mixture of  fluorene and benzoic acid = 558 mg

Mass of pure fluorene  after purification = 185 mg of fluorene

Mass of pure  benzoic acid after purification = 144 mg of benzoic acid

Percentage of fluorene int he mixture = 185mg/558mg * 100 = 33.15 %

Percentage of benzoic acid in the mixture = 144mg/558mg * 100 = 25.81 %

Percentage of impurities = 100% - [33.15 + 25.81]

Percentage of impurities = 100 - 58.96

Percentage of impurities = 41.04 %

PLEASE HELP ASAP I NEED IT HELP PEASEEEEEEEEEEE it’s number 4 (this is science)

Answers

Here you go:
Continent Polar - Cold and Dry
Maritime Polar - Cold and Humid
Arctic - Extremely Cold and Dry
Maritime Tropical- Warm and Humid

Hope this helps:)

A student is given a sample of an unknown solid and asked to identify it. The student begins by making a list of questions to investigate about the sample.
BRAINLY QUICK!

Which question would be the least useful in helping the student identify the sample?


A) Is the sample soluble in water?

B) What is the weight of the sample?

C) Is the sample attracted to a magnet?

D) What is the melting point of the sample?

Answers

Answer:

B) What is the weight of the sample?

Explanation:

The least useful question for the student to identify the sample is "What is the weight of the sample?"

To identify a sample, it is important to know its solubility (homogenous or heterogenous), chemical properties such as melting and boiling point, and whether sample is attracted towards the magnet or not to know the nature of the sample.

Weights is the least useful characteristic of a sample that can help to identify the sample.

Hence, the correct answer is "B) What is the weight of the sample?".

The weight loss of an aluminum alloy corroding in a solution of hydrochloric acid was observed to be 0.250 g/cm2 after an 8 h immersion period [13]. What is the corresponding anodic current density in milliamperes per square centimeter, assuming that all the corrosion is due to the following anodic half-cell reaction:

Answers

This question is incomplete, the complete question is;

The weight loss of an aluminum alloy corroding in a solution of hydrochloric acid was observed to be 0.250 g/cm² after an 8 h immersion period [13]. What is the corresponding anodic current density in milliamperes per square centimeter, assuming that all the corrosion is due to the following anodic half-cell reaction:

Al → Al³⁺ + 3e⁻

The atomic weight is 58.7 g/mol

Answer:

Current density in milliamperes per square centimeter is 45.81 mA/cm²

Explanation:

Given the data in the question;

we know that Faradays law of electrolysis states that;

amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell.

so

m = ( Q/F) (M/z)

where m is the mass of substance liberated at an electrode( 0.250 g/cm²)

Q is the total electric charge passed through the substance

F is Faradays constant (96,500 C/mol)

M is the molar mass of the substance( 58.7 g/mol )  and Z is the number of electrons transferred(3)

Also

we know that; Q = It

where I is current and t is time( 8hrs × 60 × 60 = 28800 sec )

we substitute  Q = It  into our equation'

we have;

m = ( It/F) (M/z)

m = ItM / Fz

mFz = ItM

I = mFz / tM

so we substitute

I = [0.250 g/cm² × 96,500 C/mol × 3] / [ 28800 × 58.7 g/mol]

I = 72375 /  1690560

I = 0.04581 A/cm²

we know that; 1 ampere = 1000 milliampere

so

I = 0.04581 × 1000

I = 45.81 mA/cm²

Therefore, current density in milliamperes per square centimeter is 45.81 mA/cm²

13 Elements within the same main groups (1-8)

Answers

Answer: Valence electrons
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