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?
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?".
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)?
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.
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:
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
Do the substances that are reactants in a chemical reaction remain the same after the reaction is complete?
atomic radius(size)______ from left to right across a period on the periodic table
Answer:
Decreases
Explanation:
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.
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)
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.
Which factor can help balance Earth’s temperature?
Answer:
Clouds
Explanation:
Because I just got it right
What type of charge do ionic compounds have
Answer:neutral
Explanation:ionic compounds is between a cation and anion so they cancel eachother out
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
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