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Liono4ka [1.6K]
3 years ago
14

Plastic is used to cover the copper wire in the power codes of appliances because plastic differs from copper in _________. 1. D

estiny 2. Hardness. 3. Phrase at room temperature 4. Electrical conductivity
Chemistry
2 answers:
saul85 [17]3 years ago
5 0

Answer:

4 electrical conductivity

klio [65]3 years ago
4 0

Answer:

Bro, its so obvious. Its electrical conductivity.

Explanation:

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Hich of the following statements is true about what happens during a chemical reaction?
postnew [5]
I believe it is <span>d. the bonds of both the reactants and the products are formed.</span>
6 0
3 years ago
What is the importance of a special habitat?
vfiekz [6]
They meet all the environmental conditions an organism needs to survive
8 0
2 years ago
Silver bromide (AgBr) has a solubility of 7.07 × 10^-7 mol/L
kirill115 [55]

a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b.  Ksp AgBr = s²

c. 5 x 10⁻¹³  mol/L

<h3>Further explanation</h3>

Given

solubility AgBr = 7.07 x 10⁻⁷ mol/L

Required

The dissolution reaction

Ksp

The solubility product constant

Solution

a. dissolution reaction of AgBr

AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b. Ksp

Ksp AgBr  = [Ag⁺]  [Br⁻]

Ksp AgBr = (s) (s)

Ksp AgBr = s²

c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³  mol/L

7 0
3 years ago
Please help me :( thank you so much and if you can show work I would really appreciate it
Katyanochek1 [597]

Answer:

D

Explanation:

On the left hand side there are a total of 4 hydrogen and 2 oxygen but on the right hand side there Is only 2 hydrogen and 1 oxygen

6 0
3 years ago
A coffee-cup calorimeter initially contains 125 g water at 24.28C. Potassium bromide (10.5 g), also at 24.28C, is added to the w
iren2701 [21]

Answer:

The solution is given below

Explanation:

Heat, q= mc∆T

q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C

q=  -1619.75J

NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.

Enthalpy Change, ∆H = 1619.75 7/ 10.5 g

                                     = 154.26 J/g

No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr

                                =10.5g/119gmol-1

                                =0.088 mol

∆H= 1619.75 J/ 0.088 mol

      = 18.41 kJ/mol  

6 0
3 years ago
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