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pav-90 [236]
2 years ago
11

Aqueous aluminum bromide and solid zinc are produced by the reaction of solid aluminum and aqueous zinc bromide . write a balanc

ed chemical equation for this reaction.
Chemistry
1 answer:
Rashid [163]2 years ago
8 0
<span>3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s) The unbalanced equation is: ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s) First, count the atoms of each element on each side of the equation: Zn 1,1 Br 2,3 Al 1,1 The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting: 3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s) Now check the atom counts again for both sides: Zn 3,1 Br 6,6 Al 1,2 Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving: 3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s) Now check the atom counts again: Zn 3,3 Br 6,6 Al 2,2 And they match. So the balanced equation is: 3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)</span>
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Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimi
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Answer:

The correct answer is 28.2 %.

Explanation:

Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,  

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= 195 atm

The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100

Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,  

%VO2 = (55/195) * 100 = 28.2%

%VN2 = (90/195) * 100 = 46%

%VHe = (50/195) * 100 = 25.64%

Hence, the percent oxygen by volume present in the blend is 28.2 %.  

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A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

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Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

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