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riadik2000 [5.3K]
2 years ago
6

Name the compound formed when a copper cation has a 2+ charge combined with a bromine anion that has a 1- charge.

Chemistry
1 answer:
Rashid [163]2 years ago
7 0

Answer: The compound is called copper (II) bromide or cupric bromide

Explanation:

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Differentiate benzene and cyclohexane
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Benzene is an aromatic compound but cyclohexane is not aromatic.

Benzene is an unsaturated molecule, but cyclohexane is saturated.
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Carbon atoms in the benzene ring have sp2 hybridization where carbon atoms in the cyclohexane have sp3 hybridization.</span>
 
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How many moles of oxygen are required to react with 12 moles of FeS2?
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2 years ago
A particle with a charge of 9.40 nC is in a uniform electric field directed to the left. Another force, in addition to the elect
juin [17]

Answer:

a. Work done by the electric force = -2.85 * ×10⁻⁵ J

b. The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. The magnitude of the electric field is 33.7kV/m

Explanation:

Given.

Charge = Q = 9.40 nC

Distance = d = 9.00 cm = 0.09m

Amount of work = 7.10×10⁻⁵ J

Kinetic energy = K = 4.25×10⁻⁵ J

a. What work was done by the electric force?

This is calculated by; change in Kinetic Energy i.e. ∆KE

∆KE = ∆K2 - ∆Kæ

Where K2 = 4.25×10⁻⁵ J

The body is released at rest, so the initial velocity is 0.

So, K1 = 0

Also, total work done = W1 + W2

Where W2 = 7.10×10⁻⁵J

So, W1 + W2 = W = K2

W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵

W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵

W = -2.85 * ×10⁻⁵ J

Work done by the electric force = -2.85 * ×10⁻⁵ J

b. What is the potential of the starting point with respect to the end point?

The change in potential energy is given as

W = ∆U

W = Q|V2 - V1| where V1 = 0 because the body starts from rest

So, W = QV2

Make V the Subject of the formula

V2 = W/Q

V2 = -2.85 * ×10⁻⁵ J / 9.40 nC

V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C

V2 = −3031.9148936170212765957V

V2 = -3.03 * 10³ V

The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. What is the magnitude of the electric field?

The magnitude of the electric field is calculated as follows;

W = -Fd = -QEd

And E = V/d

E = -3.03 * 10³ V / 0.09 m

E = −33687.943262411347517730 V/m

E = -33.7kV/m

The magnitude of the electric field is 33.7kV/m

4 0
3 years ago
PLEASE HELP
HACTEHA [7]
Not sure gets hotter prolly
5 0
2 years ago
Read 2 more answers
The standard cell potential (E°cell) for the reaction below is +1.10V. The cell potential for this reaction is ________ V when t
alexandr402 [8]

Answer: 0.94 V

Explanation:

For the given chemical reaction :

Zn(s)+Cu^{2+}(aq)\rightarrow Cu(s)+Zn^{2+}

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +1.10 V

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=+1.10-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{2.5}{1.0\times 10^{-5}}

E_{cell}=+1.10-0.16V=0.94V

The cell potential for this reaction is 0.94 V

5 0
3 years ago
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