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saveliy_v [14]
3 years ago
11

Which atom has five electrons in its outer level and 10 electrons in the inner energy levels

Chemistry
1 answer:
Sever21 [200]3 years ago
6 0

Oxygen 47 Hydrogen 63

Silicon 28 Oxygen 25.5

Aluminum 7.9 Carbon 9.5

Iron 4.5 Nitrogen 1.4

Calcium 3.5 Calcium 0.31

Sodium 2.5 Phosphorus 0.22

Potassium 2.5 Chlorine 0.03

Magnesium 2.2 Potassium 0.06

Titanium 0.46 Sulfur 0.05

Hydrogen 0.22 Sodium 0.03

Carbon 0.19 Magnesium 0.01

All others <0.1 All others <0.01 Living matter

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A block is pulled 0.90 m to the right in 2.4 s. What is the block's average speed to the nearest hundredths of a m/s?
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He standard reduction potentials of lithium metal and chlorine gas are as follows: reaction reduction potential (v) li+(aq)+e−→l
AURORKA [14]

 2 Li(s)  +Cl₂→  2 Li⁺ (aq)  + 2Cl⁻ (aq)

The cell potential   of  the reaction above   is +4.40V

<em><u>calculation</u></em>

Cell  potential  =∈° red - ∈° oxidation

in  reaction above  Li  is  oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04

  Cl  is  reduce   from oxidation  state 0 to -1 therefore  the ∈°red = +1.36 V

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3 years ago
Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volum
Strike441 [17]

Answer:

35.41 L

Explanation:

Given, Volume of Copper = 4.84 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density  Volume = 8.95 g/cm³  4.84 cm³ = 43.318 g

Mass of copper = 43.318 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{43.318\ g}{63.546\ g/mol}

Moles of copper = 0.6817 moles

Given, Volume of nitric acid solution = 227 mL = 227 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density  Volume = 1.42 g/cm³  227 cm³ = 322.34 g

Also, Nitric acid is 68.0 % by mass. So,  

Mass of nitric acid = \frac {68}{100}\times 322.34\ g = 219.1912 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{219.1912\ g}{63.01\ g/mol}

Moles of nitric acid = 3.4786 moles

According to the reaction,  

Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}

1 mole of copper react with 4 moles of nitric acid

Thus,  

0.6817 moles of copper react with 4*0.6817 moles of nitric acid

Moles of nitric acid required = 2.7268 moles

Available moles of nitric acid = 3.4786 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.6817 mole of copper on reaction forms 2*0.6817 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.3634 moles

Given:  

Pressure = 724 torr

The conversion of P(torr) to P(atm) is shown below:

P(torr)=\frac {1}{760}\times P(atm)

So,  

Pressure = 724 / 760 atm = 0.9526 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.35 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9526 atm × V = 1.3634 mol × 0.0821 L.atm/K.mol × 301.35 K  

⇒V = 35.41 L

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Answer:

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Explanation:

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