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timurjin [86]
3 years ago
10

Please please help:(

Chemistry
1 answer:
borishaifa [10]3 years ago
4 0
Magnesium and silver
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2) In a calorimetry experiment, 70.0 g of a substance was heated to
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The temperature at thermal  equilibrium : 16.1 °C

<h3>Further explanation</h3>

Q absorbed = Q released

Heat can be formulated :

Q = m.c.Δt

c= specific heat, J/g°C

Δt = temperature different, C°

m= mass, g

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\tt 3500\times 4.18\times (t_2-15)=70\times 0.89\times (270-t_2)\\\\14630(t_2-15)=62.3(270-t_2)\\\\14630t_2-219450=16821-62.3t_2\\\\14692.3t_2=236271\\\\t_2=16.1^oC

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4.14 Oxygen requirement for growth on glycerol Klebsiella aerogenes is produced from glycerol in aerobic culture with ammonia as
Iteru [2.4K]

Answer:

0.37 g

Explanation:

The molecular weight for Glycerol = 92

Number of Carbon atoms in glycerol (x)  C_3H_8O_3 = 3

Molecular weight of  the Biomass  ( Klebsiella aerogenes )

= CH_{1.73}O_{0.43}N_{0.24}

= \frac{23.97}{0.92}

= 26.1

From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as \delta _g) as follows:

= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)

= 4.15

Given that the yield of the Biomass = 0.40 g

However;

C = Yield of Biomass *\frac{Molecular weight of substrate}{Molecular weight of the Biomass}

C = 0.40*\frac{92}{26.1}

C = 1.41 g

Now , the oxygen requirement can be calculated as:

= \frac{1}{4}*(n*S -  C * \delta _{g})

= \frac{1}{4}(3*4.7-1.41*4.15)

= 2.1 g/mol

Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.

Now converting it to mass terms; we have:

= 2.1*\frac{number of mole of oxygen}{molecular weight of glycerol}

= 2.1 * \frac{16}{92}

= 0.3652 g

≅ 0.37 g

∴ The oxygen requirement for this culture in mass terms = 0.37 g

3 0
3 years ago
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