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3 years ago

Classify the chemical reaction: 2Al + 3NiSO4 ⟶ Al2(SO4)3 + 3Ni PLEASE HELP!

2 answers:

8 0

Following are the types of chemical reactions with a brief overview:

<span>1) Combination (Two elements combine to form a compound)
2) Decomposition (A compound is broken into elements)
3) Single displacement (An element displaces an element present in a compound)
4) Double displacement (The elements of two compounds displace each other)
5) Combustion (Burning Takes places)
<span>6) Redox (Oxidation Reduction Reaction)

If you see the given equation, Al is displacing Ni in the compound NiSO4 and taking its place. Ni is set free as a separate element. So this type of reaction is known as Single Displacement Reaction.</span></span>

user100 [1]3 years ago

3 0

Kc = [Alsmall2(SOsmall4)small3] [Ni]^3 / [Al]^2[NiSOsmall4]^3

I hope this help.
And how are you doing chemistry if your I'm middle school?

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Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

a chemical reaction produced 2.50 moles of nitrogen gas. what volume in liters, does this gas sample occupy at STP? (show your w

Harrizon [31]

The volume of N₂ at STP=56 L

<h3>Further explanation</h3>

Given

2.5 moles of N₂

Required

The volume of the gas

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.

So for 2.5 moles gas :

$\tt 2.5\times 22.4=56~L$

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3 years ago

Calculate mass of hydrogen chloride produced from 11.3g H and 11.3g Cl. <br> H2 + Cl2 -> 2HCl

S_A_V [24]

Answer:

22.6According to the law of conservation of mass, the total mass of reactants should be equal to the total mass of products. Here, total mass of reactants= 11.3+11.3=22.6

•

• • Mass of Hcl produced=22.6 g

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The average kinetic energy of a body is called

Veronika [31]

Temperature is a measure of average kinetic energy of particles in an object. The hotter the substance, higher is the average kinetic energy of its constituent particles.

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The common pathway for the oxidation of glucose and fatty acids is ________.

finlep [7]

The citric acid cycle should be the answer.

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