CHEMISTRY COLLEGE

Identify each element below, and give the symbols of the other elements in its group. a. [Ar] 4s23d104p4
b. [Xe] 6s24f145d2
c. [Ar] 4s23d5.

Answers

Answer 1
Answer:

Answer:

Explanation:

the electron configuration is defined as the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It is

used to describe the orbitals of an atom in its ground state

The valence electrons, electrons in the outermost shell, can be used to know the chemical property

a)

Chemical Name of the Element: Selenium

Chemical Symbol: Se

Group it belong in periodic table:6A

Other Element in the same group:tellurium(Te),,sulfur(S)

atomic number = 34

Selenium is a chemical element that has symbol Se It is a nonmetal which is usually classified as metalloid with properties that are intermediate between the elements above and below in the periodic table.

b)Chemical Name of the Element:Hafnium

Chemical Symbol: Hf

Group it belong in periodic table:4B

Other Element in the same group: Titanium( Ti )Rutherfordium

atomic number: 72

Hafnium is a solid at room temperature.

c)Chemical Name of the Element: Manganese

Chemical Symbol:Mg

Group it belong in periodic table:Mn

Other Element in the same group:Bohrium(Bh) ,Technetium(Tc)

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How much energy (in J) is lost when a sample of iron with a mass of 26.4 g cools from 74.0 ∘C to 26.0 ∘C?

Answers

Answer:

Q=-526.6J

Explanation:

Hello,

In this case, for the computation of the energy loss when the cooling process is carried out, we use the shown below equation:

Q=mCp\Delta T

Whereas we need the mass, specific heat and change in temperature of iron within the process. Thus, the only value we need is the specific heat that is 0.444 J/(g°C), therefore, we compute the heat loss:

Q=26.4g*0.444(J)/(g\°C)*(26.0\°C-74.0\°C)\n \nQ=-526.6J

Negative sign points out the loss due to the cooling.

Regards.

1. What is the angular distance north or south from the earth's equator beginning at0° at the equator and ending at 90° at either pole?
a. Weather
b. Altitude
c. Latitude
d. Climate

Answers

Answer:

c. Latitude

Explanation:

The angular distance north or south from the earth's equator beginning at 0° at the equator and ending at 90° of either pole is the latitude.

The equator is a line of latitude that divided the earth into two hemispheres.

Only the equator is a great circle as a line of latitude. Others are small circles.

  • Weather is the atmospheric condition of a place over a short period of time
  • Climate is the average weather condition at a place over a long period of time.
  • Altitude of a place is its elevation above a reference plane.

A solution containing 292 g of Mg(NO3)2 per liter has a density of 1.108 g/mL. The molality of the solution is:A) 2.00 m

B) 1.77 m

C) 6.39 m

D) 2.41 m

E) none of these

Answers

Answer: D) 2.41 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molality=(n)/(W_s)

where,

n = moles of solute

 W_s = weight of solvent in kg

moles of solute =\frac{\text {given mass}}{\text {molar mass}}=(292g)/(148g/mol)=1.97moles

volume of solution = 1L = 1000 ml      (1L=1000ml)

Mass of solution={\text {Density of solution}}* {\text {Volume of solution}}=1.108g/ml* 1000ml=1108g

mass of solute = 292 g

mass of solvent = mass of solution - mass of solute = (1108- 292) g = 816g = 0.816 kg

Now put all the given values in the formula of molality, we get

Molality=(1.97moles)/(0.816kg)=2.41mole/kg

Therefore, the molality of solution will be 2.41 mole/kg

Final answer:

In this problem, we calculate molality by using the given mass of the solute, the mass of the solvent, and the molar mass of the solute. After performing the necessary calculations, we find that the molality is 2.41 m.

Explanation:

The subject of this student's question is molality, which is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. To find the molality (m), we need to know the mass of the solute and the mass of the solvent in the solution.

Given, that the solution contains 292g of Mg(NO3)2 per liter (which is the mass of the solute). The density of the solution is 1.108g/mL. We know that 1L = 1000mL, so the mass of the solution is density x volume = 1.108g/mL x 1000mL = 1108g.

We need to find the mass of the solvent (water). The mass of the solution is the mass of the solute + the mass of the solvent. So, the mass of the solvent is 1108g(mass of the solution) - 292g(mass of solute) = 816g or 0.816gkg.

The molar mass of Mg(NO3)2 is 148.31452 g/mol. So, the number of moles of Mg(NO3)2 in the solution is moles = mass / molar mass = 292g / 148.31452 g/mol = 1.97 moles.

Now we can calculate molality (m) = moles of solute/mass of solvent in kg = 1.97 moles / 0.816 kg = 2.41 m. Therefore, the answer is D) 2.41 m.

Learn more about Molality here:

brainly.com/question/26921570

#SPJ3

Consider the combination reaction of samarium metal and oxygen gas. If you start with 33.7 moles of samarium metal, how many moles of oxygen gas would be required to react completely with all of the samarium metal? For this reaction, samarium has a +3 oxidation state within the samarium/oxygen compound.

Answers

Answer:

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

4Sm+3O_2\rightarrow 2Sm_2O_3

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of  samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

(3)/(4)* 33.7 mol=25.275 molof oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Answer:

Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is \fbox{25.3 \text{ mol}}.

Explanation:

A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.

The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:

1. Amount of one reactant required to react completely with the other reactant.

2. Amount of the product that can be produced from the given amount of the reactant.

Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.

The chemical formula for oxygen gas is \text{O}_(2).

Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is \text{Sm}_(2)\text{O}_(3).

The chemical equation is as follows:

\fbox{\text{Sm}+\text{O}_(2) \rightarrow \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}

Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.

The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of \text{Sm}_(2)\text{O}_(3) and 3 in front of \text{O}_(2) to balance the oxygen atoms.

\fbox{\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}

The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.

\fbox{\n4\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}

Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.

According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.

Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:

\text{moles of O}_(2) = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)                               ...... (1)

Step 5: Substitute 33.7 mol for moles of Sm in equation (1).

\text{Moles of O}_(2) = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)\n\text{Moles of O}_(2)= 25.275 \text{ mol}\n\text{Moles of O}_(2)= 25.3 \text{ mol}

Note:

Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Some basic concept of chemistry

Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.

QUESTION 2: The molar mass for this compound is 1.11g / m * o * l The molecular formula for this compound is

Answers

I need the percentage again dude

How many molecules are in 2.50 moles of co2

Answers

Explanation:

There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.
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