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Vera_Pavlovna [14]
2 years ago
13

If a zero order reaction has a rate constant k of 0.0416Mmin and an initial concentration of 2.29 M, what will be its concentrat

ion after 20.0 minutes? Your answer should have three significant figures.
Chemistry
1 answer:
babymother [125]2 years ago
7 0

Answer:

The concentration after 20 mins is 0.832 M

Explanation:

Zero order rate law is given by;

R = K [A₀]⁰

A zero order reaction, rate is independent of the initial concentration

R = K

Where;

R is the rate of reaction

K is the rate constant = 0.0416 M/min

Since R = K,

Then, R = 0.0416 M/min

After 20 min, the concentration will be;

A = Rt

A = (0.0416 M/min)(20 min)

A = 0.832 M

Therefore, the concentration after 20 mins is 0.832 M

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Determine the number of moles and mass requested for each reaction in Exercise 4.42.
suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(c) 8 mol NaNO₃ and 680g NaNO₃

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

  1. Step 1: Write the balanced equation
  2. Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
  3. Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. <u>46g of Na react with 1 mol of Cl₂</u>. Since the molar mass of Cl₂ is 71g/mol, then <u>46g of Na react with 71g of Cl₂</u>.

Step 3:

10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}

10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

<u>216.5g of HgO</u> form <u>0.5 moles of O₂</u>. <u>216.5g of HgO</u> form <u>16g of O₂</u>.

Step 3:

1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}

1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

<u>16g of O₂</u> come from <u>1 mol of NaNO₃</u>. <u>16g of O₂</u> come from <u>85g of NaNO₃</u>.

Step 3:

128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}

128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

<u>12 g of C</u> form <u>1 mol of CO₂</u>. <u>12 g of C</u> form <u>44g of CO₂</u>.

Step 3:

20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2

[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2[/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

<u>79.5g of CuO</u> come from <u>1 mol of CuCO₃</u>. <u>79.5g of CuO</u> come from <u>123.5g of CuCO₃</u>.

Step 3:

1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}

5 0
2 years ago
How many moles of O2 should be supplied to burn 1 mol of C3H8 (propane) molecules in a camping stove
Shkiper50 [21]
The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Each mole of propane requires 5 moles of oxygen.
4 0
1 year ago
The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi
BARSIC [14]
The answer is: D) centi
5 0
2 years ago
Annie is trying to attract her metal pencil box with a magnet from the other side of her desk. The pencil box does not move. Wha
elena55 [62]

Answer:D

Explanation:because The farther an object is from a magnet are apart from each other, the weaker the repulsion force will be.

8 0
2 years ago
1. For which of these elements would the first ionization energy of the atom be higher than that of the diatomic molecule?
Lyrx [107]

Answer: Option (b) is the correct answer.

Explanation:

The energy necessary to remove an electron from a gaseous atom or ion is known as ionization energy.

This means that smaller is the size of an atom more amount of energy has to be supplied to it in order to remove the valence electron. This is because in small atom or element there will be strong force of attraction between the nucleus and electrons.

So, high amount of energy has to be supplied to remove the valence electrons.

As electronic configuration of helium is 1s^{2}. So, due to completely filled valence shell it is more stable in nature.

As a result, we need to provide very high amount of energy to remove an electron from a helium atom.

Thus, we can conclude that out of the given options helium element would the first ionization energy of the atom be higher than that of the diatomic molecule.

7 0
2 years ago
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