A(n) __________ is a coil of wire that produces a magnetic field when it carries an electric current

The initial reaction rate for the elementary reaction 2A + B → 4C was measured as a function of temperature when the concentrati

Artist 52 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The activation energy is 124.776 $\frac{kJ}{mole}$

b) The frequency factor is 1.77 × $10^{18}$

c) The rate constant is 0.00033 $(\frac{dm^{3} }{mole} )^{2}$ $\frac{1}{s}$

Explanation:

From the question the elementary reaction for A and B is given as

2A + B → 4C

The rate equation the elementary reaction is

- $r_{A}$ = $k$ $[A]^{2}$ $[B]$

= $k[2]^{2}[1.5]$

= 6k

$k = \frac{-r_{A} }{6}$

When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.

The relationship between temperature and rate constant can be deduced from these equation

$k = Aexp(-\frac{E_{a} }{RT} )$

taking ln of both sides we have

$lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}$

Considering the graph for the rate constant $ln k$ and $(\frac{1}{T} )$ the slope from the equation is $-(\frac{E_{a} }{R})$ and the intercept is $ln A$

From the given table we can generate another table using the equation above as shown on the third uploaded image

The graph of $ln k$ vs $(\frac{1}{T} )$ is shown on the fourth uploaded image

From the graph we can see that the slope is $-(\frac{E_{a} }{R} ) = - 15008$

Now we can obtain the activation energy $E_{a}$ by making it the subject in the equation also generally R which is the gas constant is $8.145 \frac{J}{kmole}$

$E_{a} = 15008 × 8,3145\frac{J}{molK}$

$= 124\frac{KJ}{mole}$

Hence the activation energy is $= 124\frac{KJ}{mole}$

b) From the graph its intercept is $ln A = 42.019$

$A = exp(42.019)$

$=1.77 × 10^{18}$

Hence the frquency factor A is $=1.77 × 10^{18}$

c) From the equation of rate constant

$lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}$

We have

$ln k = 42.019 - 15008 * (\frac{1}{300} )$

$k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}$

Hence the rate constant is $k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}$

6 0

3 years ago

Concerning the 10.0 mL of 0.50 M NaCl to 100 mL of solution: When a solution is diluted, does it change the number of moles diss

Mars2501 [29]

No, the dilution does not change the number of moles dissolved

Explanation:

We can see that,

The molarity of the solution was 0.50 M

The volume of the solution is 10 ml.

No of moles of the solute was= volume * concentration

= 10 X 10^-3* 0.50

= 5*10^-3 moles

When the solution is diluted from 10 ml to 100ml, the molarity or concentration changes but number of moles remains constant.

The molarity of 100 ml solution will be

c=n/V

= 5*10^-3*/100*10^-3

= 0.05

when the solution is diluted to 100ml from 10 ml molarity changes from 0.5M TO 0.05 M

8 0

3 years ago

Read 2 more answers

The Following questions pertain to a 2.2M solution of hydrocyanic acid at 25°C. pKa = 9.21 at 25°C. Find the concentrations of a

horsena [70]

1) Chemical reaction: HCN + H₂O → CN⁻ + H₃O⁺.
c(HCN) = 2,2 M = 2,2 mol/L.
pKa(HCN) = 9,21.
Ka = 6,16·10⁻¹⁰.
[CN⁻] = [H₃O⁺] = x.
[HCN<span>] = 2,2 M - x.
</span>Ka = [CN⁻] · [H₃O⁺] / [HCN].
6,16·10⁻¹⁰ = x² / 2,2 M -x.
Solve quadratic equation: [CN⁻] = [H₃O⁺] = 0,0000346 M.
[HCN] = 2,2 M - 0,0000346 M = 2,199 M.

2) pH = - log[H₃O⁺].
pH = -log( 0,0000346 M).
pH = 4,46.
Hydrocyanic acid and hydronium ion (H₃O⁺) are acids. Cyanide anion (CN⁻) is the strongest base in the system, cyanide anion accept protons in chemical reaction.
pKb = pKw - pKa.
pKb = 14 - 9,21 = 4,79.

6 0

3 years ago

How do the muscular and skeletal systems work together to allow us to move?

BlackZzzverrR [31]

I would say it is B because when u pull ur leg up the space in your muscles are smaller when you spread your leg out they relax because they are not pulling anything.<span />

7 0

3 years ago

H2O is polar covalent or its Hydrogen bond??? PLEASE HELP!!!

Annette [7]

Answer:

Polar covalent.

Explanation:

The covalent bonds are therefore polar, and the oxygen atoms have a slight negative charge (from the presence extra electron share), while the hydrogens are slightly positive (from the extra un-neutralized protons). Opposite charges attract one another.

6 0

3 years ago