A tetrahedron is like a pyramid, but with a triangular base. How might atoms bond to form this shape?
1 answer:
Tetrahedron generally known as triangular pyramid consists of 4 triangular faces, 6 edges and 4 vertices. The base of the tetrahedron is a triangle. All the vertices have the same atom that is bonded to the central metal atom which lies in the centre of the tetrahedron. We take as an example to see how the atoms actually bond in tetrahedron. It is quite visible from the image that 4 O-atoms lies on all the 4 vertices and 1 Si-atom lie in the centre.
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Answer:
1.37 x CFU/mL
Explanation:
First, the dilution factor needs to be calculated.
Since four 9 ml dilution blanks were prepared, the dilution factor that yielded 137 colonies is of .
Next is to divide the colony forming unit from the dilution by the dilution factor:
137/ = 137 x
In order to get the CFU/ml, divide the CFU from the dilution by the plated volume (1 mL) from the final dilution tube.
137 x /1 = 1.37 x
Hence, the CFU/ml present in the original <em>E. coli </em> sample is 1.37 x .
cfu/ml = (no. of colonies x dilution factor) / volume of culture plate
Answer:
1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Explanation:
<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>
example :
Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )
Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment
Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )
Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
<u>2) Entropy change for Decomposition of mercuric oxide </u>
2HgO (s) → 2Hg(l) + O₂ (g)
Δs = positive
there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C
hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =
Δh of reaction = 181.6 KJ
Temp = 500 + 273 = 773 k
hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Answer:
Therefore, the amount of heat produced by the reaction of 42.8 g S = <u>(-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>
Explanation:
Given reaction: 2S + 3O₂ → 2 SO₃
Given: The enthalpy of reaction: ΔH = - 792 kJ
Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol
In the given reaction, the number of moles of S reacting: n = 2
As, Number of moles:
∴ mass of S in 2 moles of S:
<em>Given reaction</em>: 2S + 3O₂ → 2 SO₃
<em>In this reaction, the limiting reagent is S</em>
⇒ 2 moles S produces (- 792 kJ) heat.
or, 64 g of S produces (- 792 kJ) heat.
∴ 42.8 g of S produces (x) amount of heat
⇒ <u><em>The amount of heat produced by 42.8 g S:</em></u>
<u>Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>