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2 years ago

Using the periodic table to locate each element, write the electron configuration of(c) Re.

1 answer:

5 0

Rhenium is a chemical element with the symbol Re and atomic number 75. The electron configuration of Re is [Xe] $6s^{2} 4f^{14} 5d^{5}$ .

<h3>How to write an electronic configuration?</h3>

1. Identify the given element and its atomic number from the periodic table.

2. Write the electron configuration by the energy level and the type of orbital first, then the number of electrons present in the orbital as superscript.

The easiest way to write the electronic configuration for any element is by using a diagonal rule for electron filling order in the different subshells according to the Aufbau principle.

The 3 rules for writing the electron configuration in the orbital box diagram are – the Aufbau rule, the Pauli-exclusion rule, and Hund's Rule.

To learn more about electronic configuration, refer

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For the following problem convert both the reactants to moles and balance chemical equationsThe reaction of 167 g Fe2O3 with 85.

saul85 [17]

Let's start by balancing the reaction:

$Fe_2O_3+CO\longrightarrow Fe+CO_2$

As we can see, C appears only on two comopunds, CO and CO₂, and since both have 1 C each, their coefficients have to be the same for C to be balanced. However, CO has 1 O and CO₂ has 2, so there is a difference of 1 O betwee them.

The other source of O is Fe₂O₃, that has 3 O. So, we must choose a coefficient for CO and CO₂ such that the difference between the numbers of O is a multiple of 3, that way we can fix this difference with the O from Fe₂O₃. So, we can put coefficients of 3 on both of them:

$Fe_2O_3+3CO\longrightarrow Fe+3CO_2$

That way, we maintained C balanced (3 on each side) and now we have 3 + 3 O on the left side and 6 O on the right side, so the same amount.

Now, we just have to calance Fe, but it is easy since we have it alone in Fe. Since we have 2 on the left side, it is enough to put a coefficient of 2 on Fe to get the balanced reaction:

$Fe_2O_3+3CO\longrightarrow2Fe+3CO_2$

Now, to convert from mass to number of moles, we need the molar masses of the reactants, which we can calculate from the atomic weights of the elemnts in each of them:

$M_{Fe_2O_3}=2\cdot M_{Fe}+3\cdot M_O=(2\cdot55.845+3\cdot15.9994)g/mol=159.6882g/mol$ $M_{CO}=1\cdot M_C+1\cdot M_O=(1\cdot12.0107+1\cdot15.9994)g/mol=28.0101g/mol$

Now, we can convert their masses to number of moles:

$\begin{gathered} M_{Fe_{2}O_{3}}=\frac{m_{Fe_2O_3}}{n_{Fe_{2}O_{3}}} \\ n_{Fe_2O_3}=\frac{m_{Fe_2O_3}}{M_{Fe_{2}O_{3}}}=\frac{167g}{159.6882g/mol}=1.045787\ldots mol \end{gathered}$ $\begin{gathered} M_{CO}=\frac{m_{CO}}{n_{CO}} \\ n_{CO}=\frac{m_{CO}}{M_{CO}}=\frac{85.8g}{28.0101g/mol}=3.063180\ldots mol \end{gathered}$

Now, to determine the limiting reactant, we need to divide both the number of mole by their coefficients on the balanced reaction, so we can see how many we need per reaction of each:

$\begin{gathered} Fe_2O_3\colon\frac{n_{Fe_2O_3}}{1}=\frac{1.045787\ldots mol}{1}=1.045787\ldots mol \\ CO\colon\frac{n_{CO}}{3}=\frac{3.063180\ldots mol}{3}=1.021060\ldots mol \end{gathered}$

Now, the limiting reactant is the one we have less number of moles per reaction. We can see that we have less CO than Fe₂O₃, so the limiting reactant is CO.

4 0

1 year ago

When heating a solution to boiling on a hot plate, start by ___________. Then, turn the heat to ___________to start. If necessar

Aleonysh [2.5K]

Answer:

1. starting and stabilizing the stir function

2. a medium heat

3.turn up the heat setting

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

Some of the laboratory apparatus (equipment) used for conducting a chemical reaction are conical flask, Bunsen burner, beaker, tongs, crucible, round bottom flask etc.

When heating a solution to boiling on a hot plate, start by starting and stabilizing the stir function. Then, turn the heat to a medium heat to start. If necessary, turn up the heat setting after waiting about ten minutes without seeing boiling.

The safety precautions that must be taken when heating a solution to boiling on a hot plate;

I. A proper inspection of the round bottom flask for cracks, irregularities or any imperfection.

II. Ensure you avoid heating the flask while it is closed.

III. When suspending the flask on a hot plate, you should ensure that you use a clamp for stability.

5 0

3 years ago

When copper oxide is reacted with sulfuric acid, the acid is often gently heated. This is done in order to:

pav-90 [236]

Copper oxide(solid) + Sulphuric Acid (aqueous)-> Copper Sulphate (aqueous)+ Water(liquid)

In equation form:

CuO +H2SO4 -> CuSO4 + H2O

The colour change you will see is black to blue as Copper oxide is usually found as a black powder. Upon the reaction with sulphuric acid it will change to a cyan blue.

If you heat the made solution of copper sulphate, the water will evaporate and you will be left with white anhydrous copper sulphate crystals.

7 0

3 years ago

Which of the following elements has the highest electronegativity?

Nastasia [14]

Answer: is c

Explanation:

8 0

3 years ago

Read 2 more answers

A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is

aliina [53]

Answer:

$\large \boxed{\text{67 000 g}}$

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

$q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}$

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

$\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\$

$\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}$

7 0

3 years ago