Answer:
Magnesium (Mg) and Barium (Ba) are most similar in terms of properties.
Explanation:
The periodic table of elements is organized with elements classified by atomic number. Each column in the table represents elements grouped together by protons in the nucleus and as a result of that also charge. The first column of elements have a +1 charge, while the second column have a +2 charge.
Electronegativity, mass and weight can also be determined by the location of elements in the table, with elements becoming more electronegative from left (least amount) to right (most) and more heavy from top to bottom. Because Magnesium and Barium are in the same column, while gold is in a different column (and assigned a different color) it indicates the element gold is not like the other two and has different properties.
Based on the organization of elements in the periodic table, the two elements that will be similar are magnesium (Mg), barium (Ba).
The periodic table is arranged in rows called periods and columns called groups.
Elements in the same group of the periodic table are chemically similar because they have the same number of electrons in their outermost shell and consequently undergo the same kind of chemical reactions.
Having said that, Barium and magnesium both belong to group 2 of the periodic table hence they have the similar properties because they are all members of the same group.
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A chemist determines by measurements that .060 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates. Be sure your answer has the correct number of significant digits.
Answer:
1.7 g N₂
Explanation:
In order to convert moles of nitrogen gas (N₂) to grams, we'll use its molar mass:
Molar mass of N₂ = (Atomic Mass of N)*2 = (14.00 g/mol)*2 = 28.00 g/mol0.060 mol N₂ * 28.00 g/mol = 1.68 g N₂0.060 only has two significant digits, so the reported answer should have two significant digits as well. Meaning the answer is 1.7 g.
PLEASE HELP ASAP I NEED IT HELP PEASEEEEEEEEEEE it’s number 4 (this is science)
Pls help, science 8th grade question, #11 thx :)
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.
Do the substances that are reactants in a chemical reaction remain the same after the reaction is complete?
Which conclusion was a direct result of the gold foil experiment?
To see the number of atoms of an element in a given molecule we need to multiply stoichiometry to the number that is written on the foot of the element that is stoichiometry. Therefore, each atom in the foil was composed mostly of empty space because most alpha particles directed at the foil.
What is atom?Atom is the smallest particle of any element, molecule or compound. Atom can not be further divided. Atoms contains nucleus in its center and electron that revolve around the atom in fixed orbit.
In the nucleus, proton and neutron are present. Electron has -1 charge while proton has +1 charge. Neutron is neutral that is it has no charge. So overall the charge of nucleus is due to only proton, not by neutron.
Each atom in the foil was composed mostly of empty space because most alpha particles directed at the foil. This conclusion was the direct result of the gold foil experiment.
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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
What type of charge do ionic compounds have
Answer:neutral
Explanation:ionic compounds is between a cation and anion so they cancel eachother out
Cellular respiration is? (Single choice.)
A. a chemical reaction that converts energy from food molecules by way of oxygen into a usable energy called ATP.
B. a chemical reaction that converts carbon dioxide and water into glucose and oxygen.
C. a chemical reaction that converts that occurs only in animals.
D. a chemical reaction that occurs only in plants.
Which factor can help balance Earth’s temperature?
Answer:
Clouds
Explanation:
Because I just got it right
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.
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?".
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
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 %
atomic radius(size)______ from left to right across a period on the periodic table
Answer:
Decreases
Explanation: