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IrinaVladis [17]
3 years ago
8

Which of these substances will conduct electricity? CuSO4(s) CuSO4(aq) C5H12(l) CO2(g)

Chemistry
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer:

CuSO4(aq)

Explanation:

The aqueous for of CuSO₄ is an electrolyte and it will conduct electricity.

CuSO₄ is an ionic substance and in solid state, it will not have the ability to conduct electricity.

  • In aqueous form, it will contain free mobile ions which serves as carriers of electric current.
  • In the solid state, the ions are arranged into a crystal lattice and will not conduct a current of electricity.
  • The remaining given compounds will not conduct electricity as they are non-polar and do not have free mobile electrons.
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The volume of a metal rod was determined to be 38.6 mL. Its
True [87]

On the basis of significant figures, the accuracy of measurement made using an analytical balance is more reliable.

<h3>What is an analytical balance?</h3>

An analytical balance is a special type of weighing balance that has a reputation for obtaining the mass of very small objects even as low as milligrams. It consists of a transparent glass casing and a scale.

On the basis of significant figures, the use of an analytical balance to determine the weight will improve the accuracy of the determination since it determines the mass to more significant figures.

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3 0
2 years ago
Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration
elixir [45]

Answer:

8.54

Explanation:

At equivalence point :  

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052             = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa  

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :  

pH = 1/2 [ pKw + pKa + log c ]  

Ka for acetic acid = 1.75 X 10⁻⁵  

so pKa = -log (1.75 X 10⁻⁵) = 4.74  

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169  

so log c = log 0.02169 = -1.66  

putting the values....  

pH = 1/2 [14 + 4.74 - 1.66 ]  

pH = 1/2 [ 17.08] = 8.54

6 0
3 years ago
Read 2 more answers
What is the molar concentration of a 2 liter solution containing 200 grams of glucose?
NARA [144]

The molar concentration is 1.11M.

<h3>What is molar concentration?</h3>

The phrase "molar concentration" (also known as "molarity," "amount concentration," or "substance concentration") refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. The most frequent measure of molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.

<h3>Given : </h3>

Volume of the solution = 2L

Mass of glucose given = 200g

Concentration of glucose= ?

<h3>Formula use: </h3>

Molarity = no. of moles of solute / volume of the solution (L)

Moles of solute = given mass of solute / molar mass of the solute

<h3>Solution: </h3>

No. of moles of solute( glucose ) = 200 / 180 = 1.11 moles'

Molarity = 1.11 / 2 = 0.5555 mol L ^(-1)

Therefore, the molar concentration of glucose in the solution = 0.555 mol L ^(-1)

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What do we beileve is at the center of the milky way
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I think the answer is C because I know

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How many moles of bromine atoms are in 2.60×10^2 grams of bromine?
AlladinOne [14]
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