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yuradex [85]
2 years ago
7

What is the visual indicator that enough of a drying agent, such as anhydrous MgSO4 or CaCl2, has been added to properly dry an

organic solution?
Chemistry
1 answer:
Nina [5.8K]2 years ago
4 0

Answer:

how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?

(A) n=m/M,

n(Al)=5.4/27=0.2 moles

n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles

Number of oxygen atoms= n(O2)*Avogadro's number

=0.15*6.02*10^23=9.03*10^22 oxgyen atoms

(B)

n=m/M

n(Al)=0.6/27=0.02222 moles

n(O2)=n(Al)*3/4=0.016666 moles

m=n*M

m(O2)=0.0166666*32=0.53333 grams

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What is the mass of phosphorus that contains twice the number of atoms found in 14
yulyashka [42]

Answer:

15.5 gm

Explanation:

What is the mass of phosphorus that contains twice the number of atoms found in 14 g of iron?

[Relative atomic mass : P = 31; Fe = 56]

14 gm Fe = 14gm/ 56 gm/mole = 14 mole gm/56gm = 14/56 mole

0.25 moles

2 X 0.25 = 0.5 moles

1 mole P = 31 gm

so

0.5 moles P =31/2 =15.5 gm

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2 years ago
The type of social bonds present in modern societies that are based on difference, interdependence, and individual rights is cal
Goryan [66]

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Organic Solidarity

Explanation:

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3 years ago
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2 science questions! 20 POINTS AND WILL REWARD BRAINLIEST!
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Assume the weight of an average adult is 70. kg, and that 420. kJ of heat are evolved per mole of oxygen consumed as a result of
seraphim [82]

Answer:

The temperature difference of the body after 3 hours = 5.16 K

Explanation:

we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹

                                   or, 1.2 mole.h⁻¹

The average heat evolved by the oxidation of foodstuffs is then:

⇒          Q avg =\frac{1.2 X 420 X 10^{3} }{70} = 7.2 kj.h⁻¹.Kg⁻¹

the heat produced after 3 h would be:

                 =    7.2 kj. h⁻¹.Kg⁻¹ x 3 h

                 = 21.6 kj. kg⁻¹

                 = 21.6 x 10³ j kg⁻¹

We know Qp = Cp x ΔT

Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹

⇒ ΔT = \frac{Qp}{Cp}

⇒ ΔT = \frac{(21.6 X 10^{3} j.kg^{-1} ) }{(4.18 j k^{-1}g^{-1})   X (1000g.kg^{-1} )}

⇒ ΔT = 5.16 K

6 0
3 years ago
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