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nikklg [1K]
1 year ago
15

Does a displacement reaction take place in 'magnesium + lead nitrate'? and if so why?

Chemistry
1 answer:
cestrela7 [59]1 year ago
5 0

Answer:

Maybe or maybe not (not sure)

Explanation:

A displacement reaction is a type of reaction where one element is displaced by another from a compound.

In the case of magnesium and lead nitrate, magnesium is more reactive than lead. Therefore, it will displace lead from lead nitrate to form magnesium nitrate and lead.

The reaction can be represented as:

Mg(s) + Pb(NO3)2(aq) → Mg(NO3)2(aq) + Pb(s)


Another answer could be;

A displacement reaction does not take place in 'magnesium + lead nitrate' because magnesium is more reactive than lead.

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eduard

Answer:

a is oxidation

b is reduction

c is reduction

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What is the main difference between the Eubacteria and Archaebacteria kingdoms and the protista, fungi, plantae and Animalia kin
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The main difference between the Eubacteria and Archaebacteria kingdoms and the protista, fungi, plantae, and animal kingdoms is that Eubacteria and Archaebacteria are prokaryotes, while protista, fungi, plantae, and animal kingdoms are eukaryotes.

<h3>What is the Six Kingdoms Classification?</h3>

In the six kingdoms of classification, there are Eubacteria, Archaebacteria, Protista, Fungi, Plantae, and Animalia. The Eubacteria is a prokaryote, and its cell membrane is made up of peptidoglycan, teichoic acid, etc. Archaea are also prokaryotes, but their cell membrane composition differs from that of bacteria, so they are classified separately.

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5 0
9 months ago
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bonufazy [111]

Answer:

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Which objects are scratched?
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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
2 years ago
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