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mestny [16]
2 years ago
12

In an experiment the chemical reaction between a piece of aluminum foil and Copper(II)Chloride solution in a beaker is observed.

The Aluminum seems to disapear over time. The balanced chemical equation is given by: 2 Al + 3 CuCl ⇒3 Cu +2 AlCl3
Can you predict the outcome of the same experiment when Aluminum is replaced with a more reactive element such as Silver?

A) No reaction would occur.
B) The silver would disappear but no solid precipitate would form.
C) The silver would disappear and a silver solid precipitate would form.
D) The silver would disappear and a brownish red solid precipitate would form.

Hint: What would the new product be?

Physics
2 answers:
telo118 [61]2 years ago
8 0
When Ag(silver) reacts with CuCl2, Ag displaces Cu from CuCl2 and forms AgCl which is soluble and brownish red solid precipitate is left in the solution. 

Hence option D is correct.
Hope this helps!
bearhunter [10]2 years ago
6 0

Answer: A) No reaction would occur.

Explanation:

Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

Example: 2Al+3CuCl\rightarrow 3Cu+2AlCl_3

Thus When aluminium is treated with copper chloride ,aluminium being more reactive than copper displaces copper from its salt solution and thus produce aluminium chloride.

But when aluminum is replaced by silver, o reaction would occur as silver is less reactive than copper and hence would not be displace it from its salt solution.

Ag+CuCl\rightarrow {\text {no reaction}}

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Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume V_{1} = 2.00\ L

Pressure P_{1}= 1.00\ atm

Pressure P_{2}= 70.0\ kPa

Temperature T_{1}= 20.0°C = 293\ K

Temperature T_{2}= 7.00°C = 280\ K

We need to calculate the volume at mountains

Using  gas law

\dfrac{PV}{T}=\ Constant

For both temperature,

\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}

Put the value into the formula

\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}

V_{2}=\dfrac{101.325\times2\times280}{293\times70}

V_{2}=2.766\ litre

Hence, The volume at mountains is 2.766 L.

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2 years ago
(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the
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klasskru [66]
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Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

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Hence, the distance to the ring from the ground is 82.76m

4 0
2 years ago
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madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

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= .0837 units

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= 2.95 x .95² / 3

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moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0
2 years ago
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