Question 1 of 10
How is each element represented in the periodic table?
A. Each element is in an order based on alphabetical order.
B. Each element is listed as an abbreviation of the first letters of its
DE ME
C. Each element is represented by a one or two-letter symbol.
D. Each element is listed in its own box based on when it was
discovered​

Answers

Answer 1

Answer:

C

Explanation:

A- incorrect, not in alphabetical order

B- incorrect, symbol for salt in NA not SA

C- correct

D- incorrect, not based on discovery

Answer 2

Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.

What is Periodic Table ?

The periodic table, also known as the periodic table of the elements, is a tabular display of the chemical elements.

It is widely used in chemistry, physics, and other sciences, and is generally seen as an icon of chemistry.

Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.

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Related Questions

DATA AND CALCULATIONS: (you must show your calculations) Part I. Determination of accuracy of a graduated cylinder Calculations: Experimental Step Measurable Mass of empty graduated cylinder 47.229 g Mass of filled graduated cylinder 71.821 g Mass of water (filled – empty) g Volume of water, calculated (calculated from mass of water, using the equation “density = mass/volume”, given the fact that the density of water is exactly 1 g/mL) mL Volume of water, measured (from the reading of the scale on the graduated cylinder) 25.0 mL Percent difference between measured and calculated volumes of water [(measured-calculated)/calculated] ×100% %

Answers

Answer:

[tex]\large \boxed{2 \, \%}[/tex]

Explanation:

1. Data

Mass of graduated cylinder              =  47.229 g

Mass of graduated cylinder + water =  71.821  g

Actual volume of water                     = 25.0     mL

2. Calculations

(a) Mass of water

Mass = 71.821 g -47.229 g  = 24.592 g

(b) Volume of water

[tex]\text{Volume} = \dfrac{\text{mass}}{\text{volume }} = \dfrac{\text{24.592 g}}{\text{ 1 g/mL}} = \text{24.592 mL}[/tex]

(c) Percent Difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{Measured - Calculated}\lvert}{ \text{Calculated}} \times 100 \,\%\\\\& = & \dfrac{\lvert 25.0 - 24.492\lvert}{24.492} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.5\lvert}{24.492} \times 100 \, \%\\ \\& = & 0.02 \times 100 \, \%\\& = & \mathbf{2 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{2 \, \%} }$}[/tex]

A solution has a hydrogen ion (or hydronium ion) concentration of 1.00×10−9 M.


What is the pH of the solution?


7.5

8.0

8.5

9.0

Answers

The ph of the solution with is 9.0

Which elements cannot have more than an octet of electrons? Select all that apply
C
S
O
N
Br

Answers

Answer:

{ Carbon, Oxygen, Nitrogen }

Explanation:

Elements can only have more than an octet of electrons if they demonstrate an expanded octet. This is if they belong to groups in or beyond the third group. Why? Well these elements have d - orbitals that they can rely on to expand the number of electrons that could otherwise be limited. * Here we are focusing on main group elements, P - block elements more specifically. *

Carbon belongs to the 2 group, and thus doesn't have an empty d - orbital. Thus, it can't have more than an octet of electrons. Sulfur belongs to group 3, hence has an empty d - orbital, and can have more than an octet of electrons. Oxygen belongs to the 2 group, and thus doesn't have an empty d - orbital, so it can't have more than an octet of electrons. Same goes for Nitrogen. Bromine belongs to group 4, thus has empty d - orbitals, and can expand further than Sulfur can - it can have more than an octet of electrons.

Solution = { Carbon, Oxygen, Nitrogen }

Which image best represents the particles in liquids

Answers

This is what I would use. Hopefully this helps:)

Answer:

The 2nd Picture represents the particles in liquids.

Explanation:

A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg?

PLEASE HELP, will mark brainliest!!!

Answers

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

[tex]p_A=\chi_A\times P_T[/tex]

[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = ?

[tex]\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}[/tex]

[tex]P_{T}[/tex] = total pressure of mixture  = 525 mmHg

[tex]{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles[/tex]

[tex]{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles[/tex]

Total moles = 1.94 + 1.35 = 3.29 moles

[tex]\chi_{O_2}=\frac{1.94}{3.29}=0.59[/tex]

[tex]p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg[/tex]

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.

A gaseous mixture of O₂ and N₂ contains 37.8% nitrogen by mass, that is, in 100 g of the mixture, there are 37.8 g of N₂. The mass of O₂ in 100 g of the mixture is:

[tex]mO_2 = 100 g - 37.8 g = 62.2 g[/tex]

We will convert both masses to moles using their molar masses.

[tex]N_2: 37.8 g \times 1 mol/28.00 g = 1.35 mol\\\\O_2: 62.2 g \times 1 mol/32.00 g = 1.94 mol[/tex]

The mole fraction of O₂ is:

[tex]\chi(O_2) = \frac{nO_2}{nN_2+nO_2} = \frac{1.94mol}{1.35mol+1.94mol} = 0.590[/tex]

Given the total pressure (P) is 525 mmHg, we can calculate the partial pressure of oxygen using the following expression.

[tex]pO_2 = P \times \chi(O_2) = 525 mmHg \times 0.590 = 310 mmHg[/tex]

A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.

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Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about 39 kJ/g. How many grams of fat will you gain if you eat 10,000 kJ (about 2390 kcal) one day and do nothing but sit relaxed for 14.3 h and sleep for the other 9.70 h

Answers

Answer:

[tex]m=107.8g[/tex]

Explanation:

Hello,

In this case, for the given information, we can compute the gained grams by firstly compute the energy consumed by 14.3 h of being sit relaxed and 9.70 h  of sleeping:

[tex]E_{sit}=120\frac{J}{s}*\frac{3600s}{1h} *14.3h*\frac{1kJ}{1000J} =6177.6kJ\\\\E_{sleep}=83\frac{J}{s}*\frac{3600s}{1h} *9.70h*\frac{1kJ}{1000J} =2898.36kJ[/tex]

Then, we compute the energy used that day:

[tex]E_T=10000kJ-2898.36kJ-6177.6kJ=4203.28kJ[/tex]

Finally, the mass by considering the consumed fat:

[tex]m=\frac{4203.28kJ}{39kJ/g} \\\\m=107.8g[/tex]

Regards.

Upon the addition of water, As2O3 is converted to H3AsO3. During the titration H3AsO3 is oxidized to H3AsO4 and MnO4- is reduced to Mn2 . Write a balanced net ionic equation for the reaction

Answers

Answer:

5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)

Explanation:

Every net balanced ionic equation is composed of a union of two half equations;

The oxidation half equation (indicating electron loss) and the reduction half equation (indicating electron gain). Remember that redox reactions is a process in which electrons are lost and gained by chemical species simultaneously. One specie looses electrons in the oxidation half equation while the other specie gains electrons in the reduction half equation.

The balanced redox reaction equation shows the overall redox process and shows at a glance the total number of elect tribe lost or gained in the redox process. The overall redox reaction equation for the titration described in the question is;

5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)

Temperature on Reaction Rate Use the drop-down menus to answer the questions. Which form of the sodium bicarbonate tablet has the most surface area? As the surface area increases, what happens to the average time required for the reaction?

Answers

Answer:Crushed, decreased

Explanation:

Just got it right

will give brainliest
Explain why someone would choose to use octane in a car engine rather than octene?

Answers

Octane just prevents the air-fuel mixture from igniting before the spark plug does it. Firing the air-fuel mixture at the proper time gives you the maximum power your engine was designed to get. Using higher-octane gasoline than your engine is designed to utilize is only wasting your money.

A sample of gas has a volume of 571 mL at a pressure of 4.04 atm. The gas is compressed and now has a pressure of 7.17 atm. Predict whether the new volume is greater or less than the initial volume, and calculate the new volume. Assume temperature is constant and no gas escaped from the container.

Answers

Answer:

The new volume is less than the initial volume.

The new volume is 322mL

Explanation:

Based on Boyle's law, the pressure of a gas is inversely proportional to its volume under constant temperature. That means if the pressure of a gas is increased, the volume decrease and vice versa. The formula is:

P₁V₁ = P₂V₂

Where P is pressure and V is volume of 1, initial state and 2, final states.

In the problem, the pressure of the gas increased from 4.04atm to 7.17atm, That means the new volume is less  than initial volume because the gas is compressed occupying less volume.

Replacing in Boyle's equation:

4.04atm*571mL = 7.17atmV₂

322mL = V₂

Beeing the new volume of the compressed gas 322mL

Choose the species that is incorrectly matched with its electronic geometry.
1. BeBr2 : linear
2. CF4 : tetrahedral
3. NH3 : tetrahedral
4. H2O : tetrahedral
5. PF3 : trigonal bipyramidal

Answers

Answer:

PF3 : trigonal bipyramidal

Explanation:

PF3 has 4 domains around the central phosphorus (3 shared pairs and one lone pair of electrons), thus the electron geometry that has 4 domains is tetrahedral not  trigonal bipyramidal

From the options the specie that is incorrectly matched is ( 5 ) ; PF₃ : trigonal bipyramidal

The specie PF₃ is composed of 3 shared pairs and one unshared pair of electrons ( i.e. It has 4 domains ) as seen in the  Lewis structure of  PF₃. therefore  when writing its electronic geometry, it should expressed/written as tetrahedral and not trigonal bipyramidal.

Hence we can conclude that The specie that is incorrectly matched is PF3 : trigonal bipyramidal

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Carry out the following operations as if they were calculations of experimental results and express each answer in standard notation with the correct number of significant figures and wtih the correct units. Provide both the answer and the units.
1. 5.6792 m + .6 m + 4.33 m
2. 3.70 g - 2.9133 g
3. 4.51 cm x 3.6666 cm

Answers

Answer:

1. [tex]10.6\; \rm m[/tex] (one decimal place.)

2.[tex]0.79\; \rm g[/tex] (two decimal places.)

3. [tex]16.5\;\rm cm^2[/tex] (three significant figures.)

Explanation:

1.

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

[tex]5.6792\;\rm m[/tex] has four decimal places.[tex]0.6\; \rm m[/tex] has only one decimal place.[tex]4.33\; \rm m[/tex] has two decimal places.

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: [tex]\rm m[/tex].)

Therefore:

[tex]\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m[/tex]. (Rounded to one decimal place.)

2.

Similarly:

[tex]\rm 3.70\; \rm g[/tex] has two decimal places.[tex]2.9133\; \rm g[/tex] has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be [tex]\rm g[/tex] (same as the unit of the two inputs.)

[tex]\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g[/tex]. (Rounded to two decimal places.)

3.

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

[tex]4.51\; \rm cm[/tex] has three significant figures.[tex]3.6666\; \rm cm[/tex] has five significant figures.

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be [tex]\rm cm \cdot cm[/tex], which is occasionally written as [tex]\rm cm^2[/tex].

[tex]\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2[/tex]. (Rounded to three significant figures.)

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