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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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Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha
Stells [14]

Answer:

296.1 day.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>

5 0
2 years ago
A pharmacist–herbalist mixed 100 g lots o St. John’s wort containing the ollowing percentages o the active component hypericin:
Anna [14]

Answer:

strength of hypericin in mixture = 0.42 %

Explanation:

given data

each lot = 100 g

active component hypericin = 0.3%, 0.7%, and 0.25%

solution

we get here percent strength o hypericin in the mixture that is

Hypericin contribution lot 1 =  \frac{0.3}{100} × 100

Hypericin contribution lot 1 =  0.3 g

and

Hypericin contribution lot 2 = \frac{0.3}{100} × 100

Hypericin contribution lot 2 = 0.7 g

and

Hypericin contribution lot 3 = \frac{0.25}{100} × 100  

Hypericin contribution lot 3  = 0.25 g

so

total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g

total 300 g mixture of hypericin = 1.25 g  

so here percent strength o hypericin in mixture is

strength of hypericin in mixture = \frac{1.25}{300} × 100  

strength of hypericin in mixture = 0.42 %

5 0
1 year ago
1. Concentrated HCl is 11.7M. What is the
SSSSS [86.1K]

Answer:

The concentration is 0,2925M

Explanation:

We use the formula

C initial  x V initial = C final x V final

11,7 M x 25 ml = C final x 1000 ml

C final= (11,7 M x 25 ml)/1000 ml = 0, 2925 M

(This formula applies to liquid solutions)

6 0
2 years ago
175 mL of Cl2 gas is held in a flexible vessel at
Gnoma [55]

Answer:

V₂ = 15.6 L

Explanation:

Given data:

Initial volume = 175 mL  (0.175 L)

Initial pressure = 1 atm

Initial temperature = 273 K

Final temperature = -5°C (-5+273 = 268 K)

Final volume = ?

Final pressure = 1.16 kpa (1.16/101=0.011 atm)

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm

V₂ = 46.9 L / 3.003

V₂ = 15.6 L

7 0
2 years ago
Read 2 more answers
Element X is found in two forms: 90.0% is an isotope that has a mass of 20.0, and 10.0% is an isotope that has a mass of 22.0. W
VMariaS [17]
M_{X} = \frac{(20.0\cdot 90\%)+(22.0\cdot10\%)}{100\%} = \frac{2020}{100} = 20.2[u]

answer: B
7 0
2 years ago
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