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ryzh [129]
2 years ago
9

help??? Show that you know how to use transitional elements by rewriting the following set of sentences into a coherent paragrap

h. Career planning is essential for all high school students. Students need to understand their strengths and weaknesses. They should learn where their talents lie. They should know career opportunities for the twenty-first century. They should know which jobs require further schooling. They should know how much the job they are interested in will pay.
Chemistry
2 answers:
const2013 [10]2 years ago
6 0

<span>Career planning is very important to all the high school students.  Career planning allows them to understand what are the weaknesses and strengths in their chosen field. They have to know in what area they excel. It is important to know what the opportunities that await them in the future are. They should know which job would require more time in schooling. They have to know which job that is suitable for their interest at the same time keeps them afloat from the daily expenses, both needs and leisure.</span>

Darina [25.2K]2 years ago
4 0


Career planning is <span>essential for all high school students. They need to understand their strengths and weaknesses in order to learn where their talents lie. They should know career opportunities for the twenty-first century, their jobs require further schooling and  how much the job they are interested in will pay.</span>
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The enthalpy change for the explosion of ammonium nitrate with fuel oil is –7198 kJ for every 3 moles of NH4NO3. What is the ent
anastassius [24]

Answer:

−2399.33 kJ

Explanation:

If NH₄NO₃ reacts with fuel oil to give a ΔH of -7198 for every 3 moles of NH₄NO₃

What is the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction

∴ For every 1 mole, we will have \frac{1}{3} of the total enthaply of the 3 moles

so, to determine the 1 mole; we have:

\frac{1}{3}*(-7198kJ)

= −2399.33 kJ

∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ

7 0
2 years ago
. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture
postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

3 0
2 years ago
As an atom's nucleus gets larger the electric charges repelling the protons get larger. To compensate for this greater electric
IrinaK [193]
i think the greater the electric charge the atom decreases in size
5 0
2 years ago
Read 2 more answers
Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.
icang [17]

Answer : 3.2 X 10^{15} g/cm^{3}

Explanation :  To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = 1.66054 X 10^{-24} g

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X  1.66054 X 10^{-24} g/amu = 1.673 X 10^{-24} g

Now, Volume = 1/6πd³ as diameter is given as 1.0 X 10^{-15} m converting it to cm will require to multiply with 100

∴ Volume  = 1/6π (1.0 X 10^{-15}mX 100 cm / 1 m)^{3}

Hence, volume =  5.236 X 10^{-40} cm^{3}

Therefore, Density = mass / volume

∴ Density =  1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}

Therefore, Density will be 3.2 X 10^{15} g/cm^{3}.

6 0
2 years ago
Read 2 more answers
Which will not appear in the equilibrium constant expression for the reaction below?
n200080 [17]

Answer:

[C] carbon solid

Explanation:

Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.

5 0
2 years ago
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