What is the pH of 0.001 M HNO,?

icang [17]

Dude use google lol.

5 0

3 years ago

Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ

Whitepunk [10]

To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020

Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹

7 0

4 years ago

A certain heat engine operates between 800 K and 300 K. (a) What is the maximum efficiency of the engine? (b) Calculate the maxi

Sholpan [36]

Answer :

(a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

Explanation : Given,

Temperature of hot body $T_h$ = 800 K

Temperature of cold body $T_c$ = 300 K

(a) First we have to calculate the maximum efficiency of the engine.

Formula used for efficiency of the engine.

$\eta =1-\frac{T_c}{T_h}$

Now put all the given values in this formula, we get :

$\eta =1-\frac{300K}{800K}$

$\eta =0.625\times 100=62.5\%$

(b) Now we have to calculate the maximum work done.

Formula used :

$\eta =\frac{Q_h-Q_c}{Q_h}=\frac{w}{Q_h}$

where,

$Q_h$ = heat supplied by hot source = 1 KJ

$Q_c$ = heat supplied by hot source

w = work done = ?

Now put all the given values in this formula, we get :

$\eta =\frac{w}{Q_h}$

$0.625=\frac{w}{1KJ}$

$w=0.625KJ$

(c) Now we have to calculate the heat discharge into the cold sink.

Formula used :

$w=Q_h-Q_c$

$Q_c=Q_h-w$

$Q_c=1-0.625$

$Q_c=0.375KJ$

Therefore, (a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

4 0

4 years ago

Calculate how many grams of iron can be made from 16.5 grams of iron(III) oxide if hydrogen gas is in excess?

miss Akunina [59]

Taking into account the reaction stoichiometry, 11.54 grams of Fe are formed from 16.5 grams of iron(III) oxide if hydrogen gas is in excess.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe₂O₃: 1 mole
H₂: 3 moles
Fe: 2 moles
H₂O: 3 moles

The molar mass of the compounds is:

Fe₂O₃: 159.7 g/mole
H₂: 2 g/mole
Fe: 55.85 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Fe₂O₃: 1 mole ×159.7 g/mole= 159.7 grams
H₂: 3 moles ×2 g/mole= 6 grams
Fe: 2 moles ×55.85 g/mole= 111.7 grams
H₂O: 3 moles ×18 g/mole= 54 grams

<h3>Mass of each product formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 159.7 grams of Fe₂O₃ form 111.7 grams of Fe, 16.5 grams of Fe₂O₃ form how much mass of Fe?

$mass of Fe=\frac{16.5 grams of Fe_{2} O_{3} x111.7 grams of Fe}{159.7 grams of Fe_{2} O_{3}}$