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2 years ago

Write the electron configuration when Sodium loses one electron

Chemistry

2 answers:

Maksim231197 [3]2 years ago

3 0

Answer:

2, 8

Explanation:

Sodium Na has 11 electrons.

The electronic configuration of sodium is 2, 8, 1

Sodium loses 1 electron, the new atom has 10 electrons.

11 - 1 = 10

2, 8

lara [203]2 years ago

3 0

Answer:

Sodium cation loses one electron, so Na+: 1s(2)2s2(2)p(6)3s(1) = Na+: 1s(2)2s(2)2p(6). P: 1s(2)2s(2)2p(6)3s(2)3p(3)

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Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast

baherus [9]

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles$

$\text{Moles of} CO=\frac{260}{28}=9.3moles$

$Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)$

According to stoichiometry :

1 mole of $Fe_2O_3$ require 3 moles of $CO$

Thus 2.8 moles of $Fe_2O_3$ will require= $\frac{3}{1}\times 2.8=8.4moles$ of $CO$

Thus $Fe_2O_3$ is the limiting reagent as it limits the formation of product and $CO$ is the excess reagent.

As 1 mole of $Fe_2O_3$ give = 2 moles of $Fe$

Thus 2.8 moles of $Fe_2O_3$ give = $\frac{2}{1}\times 2.8=5.6moles$ of $Fe$

Mass of $Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g$

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

$\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%$

Therefore, the percent yield is, 91.8%

6 0

3 years ago

In order to perform a chemical reaction, 225 mL of 0.500 M lead (II) nitrate is required. How much 5.00 M stock solution is requ

vladimir1956 [14]

Answer:

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Explanation:

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4 0

2 years ago

a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp

Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH = 11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity = 4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

Learn more about heat of solution:brainly.com/question/24243878

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4 0

1 year ago

When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef

kumpel [21]

Answer:

molar mass of the phosphorus chloride = 138.06 g/mol

Explanation:

mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.

mass = 3.51 g

lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

From graham's equation of effusion,

$\frac{Rc}{Rp}$ = $\sqrt{\frac{Mp\\}{Mc} }$

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

$\frac{3.51}{1.98}$ = $\sqrt{\frac{Mp\\}{44.01} }$

1.77 = $\frac{\sqrt{Mp} }{6.64}$

11.75 = $\sqrt{Mp}$

Mp = $11.75^{2}$

Mp = molar mass of the phosphorus chloride = 138.06 g/mol

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