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Tasya [4]
1 year ago
9

When performing the last step in the reaction, aluminum metal is added in excess to the solution. It reacts and dissolves. What

happens to the excess aluminum metal when recovering the copper metal?.
Chemistry
1 answer:
alex41 [277]1 year ago
8 0

Using the law of the conservation of mass, we can see that the aluminum metal would precipitate out of the solution as aluminum oxide.

<h3>What is the last step of the reaction?</h3>

Let us recall that the law of the conservation of mass states that mass can not be created nor destroyed but the mass can be converted from one form to another. In this case, we can see that the mass of the system would remain a constant. The mass of the system does not change this is in accordance with the law of the conservation of mass.

Hence, When performing the last step in the reaction, aluminum metal is added in excess to the solution. It reacts and dissolves, there would be a recovery of the aluminum as the oxide of the aluminum and by  doing the mass of the system would remain the same but before and after the reaction.

Learn more about conservation of mass:brainly.com/question/13383562

#SPJ1

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krek1111 [17]

Answer:

Dubnium: Group 5 Period 7

Hydrogen: Group 1 Period 1

Explanation:

8 0
3 years ago
Read 2 more answers
Acetone major species present when dissolved in water (CH3)2CO
Marianna [84]

Answer:

<em>Hydrogen bonding</em>

Explanation:

<em>Acetone major species present when dissolved in water is called hydrogen boding. these occurs when, the acetone and water as the oxygen of acetone's cabonyl bond with the O-H of water.</em>

<em>Such presence of hydrogen bonding would helps the ability of molecules of two types to be miscible together</em>

4 0
3 years ago
A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron metal = 32.3 g

m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

3 0
3 years ago
A sample of gas has a mass of 827 mg . Its volume is 0.270 L at a temperature of 88 ∘ C and a pressure of 975 mmHg . Find its mo
avanturin [10]

Steps:

Mw = w * R * T / p * V

T = 88 + 273 => 361 K

p = 975 mmHg in atm :

1 atm  = 760 mmHg

975 mmg / 760 mmHg =>  1.28 atm

Therefore:

= 0.827 * 0.0821 * 361 /  1.28 * 0.270

=  24.51 / 0.3456

molar mass =  70.92 g/mol



7 0
3 years ago
In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53 × x10−
sergiy2304 [10]

the force between the electron and the proton.
a) Use F = k * q1 * q2 / d² 
where k = 8.99e9 N·m²/C² 
and q1 = -1.602e-19 C (electron) 
and q2 = 1.602e-19 C (proton) 
and d = distance between point charges = 0.53e-10 m 
The negative result indicates "attraction". 

the radial acceleration of the electron. 
b) Here, just use F = ma 
where F was found above, and 
m = mass of electron = 9.11e-31kg, if memory serves 
a = radial acceleration 

the speed of the electron. 
c) Now use a = v² / r 
where a was found above 
and r was given 

<span> the period of the circular motion.</span>
d) period T = 2π / ω = 2πr / v 
where v was found above 
and r was given 
3 0
3 years ago
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