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galina1969 [7]
3 years ago
15

The first-order decomposition of cyclopropane has a rate constant of 6.7 x 10-4 s-1. if the initial concentration of cyclopropan

e is 1.33 m, what is the concentration of cyclopropane after 644 s?
a. 0.43 m
b. 0.15 m
c. 0.94 m
d. 0.86 m
e. 0.67 m
Chemistry
2 answers:
Triss [41]3 years ago
7 0
For a first-order reaction, the equation is written below:

lnA = lnA₀ - kt
where
A₀ is the original concentration
A is the concentration left after time t
k is the rate constant

Substituting the values:
lnA = ln(1.33 M) - (6.7×10⁻⁴ s⁻¹)(644 s)
Solving for A,
A = 0.863 M

<em>Therefore, the answer is letter D.</em>
vaieri [72.5K]3 years ago
5 0
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:

ln [A] = -kt + ln [A]_o, where,

k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species

For this problem, we simply substitute the known values to the equation as in:

ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M) 

We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
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Which of the following is NOT a reason for the experimental volume of the flask to be incorrect?
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2

Explanation:

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The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x
Andrews [41]
Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
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6 0
3 years ago
A solution contains 0.140 mol KCl in 2.100 L water. What is the concentration of KCl in g/L?
viva [34]
<h3><u>Answer</u>;</h3>

≈ 4.95 g/L

<h3><u>Explanation;</u></h3>

The molar mass of KCl = 74.5 g/mole

Therefore; 0.140 moles will be equivalent to ;

 = 0.140 moles × 74.5 g/mole

 = 10.43 g

Concentration in g/L

   = mass in g/volume in L

   = 10.43/2.1

   =  4.9667

<h3>   <u> ≈ 4.95 g/L</u></h3>
4 0
3 years ago
If an element has 2+ valence electrons, does it transfer only one or more than one valence electrons​
just olya [345]

Answer:

element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.

Explanation:

  • Group IIA have 2+ valency and two electrons in its valance shell.
  • Its Electropositivity is  high and have the tendency to donate it two electrons.
  • Element  of IIA form ionic with most electronegative element.

Examples:

Cu²⁺, Mg²⁺, Sr²⁺ are examples having  2+ valance electron

one of the following is examples of element that have 2+ valence electrons

MgCl₂

Atomic number of Magnesium (Mg) is 12

Electronic Configuration of Mg:

1s², 2s², 2p⁶, 3s²

or

K =2

L = 8

M = 2

So, it have to give its 2 electrons to form a stable compound.

Similarly

Chlorine atomic number is 17

Electronic Configuration of Chlorine:

1s², 2s², 2p⁶, 3s², 3p⁵

or

K =2

L = 8

M = 7

So, it have to gain one electrons to form a stable compound and complete its octet.

So,

Two chlorine atom as a molecule gain 2 electrons from Mg²⁺ atom

So one Mg²⁺ and 2 Cl⁻ atoms form an ionic bond

where in this ionic bond Mg²⁺ transfer its 2 valence electron completely and chlorine molecule accept 2 electrons.

                 Cl-----Mg------Cl

So the Answer is

element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.

8 0
3 years ago
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