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wel
3 years ago
7

Strontium crystallizes in a face-centered cubic unit cell having an edge length of 77.43 pm. What is the atomic radius of stront

ium (in picometers) based on this structure
Chemistry
1 answer:
kiruha [24]3 years ago
5 0

Answer:

Atomic radius of Strontium is 27.38pm

Explanation:

In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:

a² + a² = b² = (4r)²

2a² = 16r²

a = √8 r

As edge length of Strontium is 77.43pm:

77.43pm / √8 = r

27.38pm = r

<h3>Atomic radius of Strontium is 27.38pm</h3>

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Answer:

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The element whose atomic number is 13 has how many valence electrons electron configuration
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It is in Group 3A = 3 valence electrons

Electron Config. = 1s2 2s2 2p6 3s2 3p1
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3 years ago
Wastewater from a cement factory contains 0.280 g of Ca2+ ion and 0.0220 g of Mg2+ ion per 100.0 L of solution. The solution den
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<u>Answer:</u> The concentration of Ca^{2+}\text{ and }Mg^{2+} ions are 2.797 ppm and 0.212 ppm respectively.

<u>Explanation:</u>

To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Volume of gold = 100 L = 100000 mL    (Conversion factor:  1 L = 1000 mL)

Density of gold = 1.001 g/mL

Putting values in above equation, we get:

1.001g/mL=\frac{\text{Mass of solution}}{100000mL}\\\\\text{Mass of solution}=1.001\times 10^5g

To calculate the concentration in ppm (by mass), we use the equation:

ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6

  • <u>Calculating the concentration of calcium ions:</u>

Mass of Ca^{2+ ions = 0.280 g

Putting values in above equation, we get:

ppm(Ca^{2+})=\frac{0.280g}{1.001\times 10^5}\times 10^6=2.797ppm

  • <u>Calculating the concentration of magnesium ions:</u>

Mass of Mg^{2+ ions = 0.0220 g

Putting values in above equation, we get:

ppm(Mg^{2+})=\frac{0.0220g}{1.001\times 10^5}\times 10^6=0.212ppm

Hence, the concentration of Ca^{2+}\text{ and }Mg^{2+} ions are 2.797 ppm and 0.212 ppm respectively.

7 0
3 years ago
Tubes through which water flows as it is brought from 0.8 MPa, 150C to 240C at essentially constant pressure in the boiler of a
motikmotik

Answer:

The correct answer is i) 50.2 % ii) 13440.906 kW and iii) 71.986 kg/s.

Explanation:

In order to find the mass flow rate of the combustion of gases, there is a need to use the energy balance equation:  

Mass of water × specific heat of water (T2 -T1)w = mass of gas × specific heat of gas (T2-T1)g

100 × 4.18 × [(240 + 273) - (150 + 273)] = mass of gas × 1.005 × [(1067+273) - (547+273)]

Mass of gas = 71.986 kg/s

The entropy generation of water can be determined by using the formula,  

(ΔS)w = mass of water × specific heat of water ln(T2/T1)w

= 100 × 4.18 ln(513/423)

= 80.6337 kW/K

Similarly the entropy generation of water will be,

(ΔS)g = mass of gas × specific heat of gas ln(T2/T1)g

= 71.986 × 1.005 ln (820/1340)

= -35.53 kW/K

The rate of energy destruction will be,  

Rate of energy destruction = To (ΔS)gen

= T₀ [(ΔS)w + (ΔS)g]

= (25+273) [80.6337-53.53)

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The availability of water will be calculated as,  

= mass of water (specific heat of water) [(T₁-T₂) -T₀ ln T₁/T₂]

= 100 × 4.8 [(513-423) - 298 ln 513/423]

= 13591.1477 kW

The availability of gas will be calculated as,  

= mass of gas (specific heat of gas) [(T₁-T₂) - T₀ ln T₁/T₂]

= 71.986 × 1.005 × [(1340-820) - 298 ln 1340/820]

= 27031.7728 kW

The exergetic efficiency can be calculated as,  

= Gain of availability / loss of availability  

= 13591.1477/27031.7728

= 0.502

The exergetic efficiency is 50.2%.  

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