1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
My name is Ann [436]
3 years ago
5

Student added a strip of aluminium metal in aqueous copper(2)Sulfate but no reaction occured. Suggest a reason why.

Chemistry
1 answer:
kari74 [83]3 years ago
7 0

Answer:

See Explanation

Explanation:

It is a common observation that a strip of aluminium metal in aqueous copper(II)Sulfate does not show any visible reaction. Aluminium is normally expected to displace copper in solution since it is higher than copper in the electrochemical series.

The reason for this is that aluminium forms an oxide film around its surface which prevents reaction with aqueous copper(II)Sulfate. This oxides film protects the aluminium surface such that it is now unable to react with the aqueous copper(II)Sulfate

You might be interested in
A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan
Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0
3 years ago
Does the amount of a substance affect the properties of a substance?
Veseljchak [2.6K]
No, the properties of a substance are not affected by the amount of a substance.
5 0
3 years ago
The second stage of photosynthesis produces _____ and _____.
lukranit [14]
The second stage of photosynthesis also called Calvin stage produces glucose

5 0
3 years ago
Read 2 more answers
Liquid does not act that a solid. how is it different from a solid and gas?
Wittaler [7]
Unlike solid matter, where particles are tightly packed and slightly vibrating, or gas, where particles go around everywhere and are extremely loose, a liquid has particles that are loosely packed but are still in slight contact with each other. Hope that's good enough
6 0
3 years ago
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
Other questions:
  • If the Earth moved farther away from the sun, which of these would probably happen?
    9·2 answers
  • A variation of the acetamidomalonate synthesis can be used to synthesize threonine. The process involves the following steps: Et
    9·1 answer
  • Since atoms have more than one electron (other than atomic hydrogen) how do you account for the electrons remaining associated w
    6·1 answer
  • 1 Which scientist first said that "elements are made up of identical atoms
    8·1 answer
  • A bag of sand and a bag of feathers can have the same density if they have different:
    12·1 answer
  • Electronegativities of the elements Na, Al, P, and Cl follow a specific trend across the period. Based on this trend, an electro
    12·2 answers
  • What is the coefficient in a chemical equation
    10·2 answers
  • Please answer this question for me I will answer your questions too I promise!!
    5·2 answers
  • Which of the following best describes a binary compound?
    5·1 answer
  • Metals have special bonds, between atoms. The valence electrons are mobile and are shared by all metal atoms present.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!