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Paraphin [41]
3 years ago
10

In terms of entropy and energy, systems in nature tend to undergo changes toward

Chemistry
1 answer:
andrew-mc [135]3 years ago
7 0
Number 2 lower entropy and higher entropy
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Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.
dybincka [34]
<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>
  1. pH:- 5.4
  2. pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

Antilog (-6)  = {10}^{ - 6} ,  \\ Antilog (0.6)  = 4

put the value in equation

{10}^{ - 6}   \times 4 = [H+] \\ 4 \times   {10}^{ - 6}  = [H+]

7 0
2 years ago
Read 2 more answers
What is the quantity of atoms in KMnO4 ?
tankabanditka [31]

Answer:

Potassium permanganate has a molar mass of 158.04 g/mol. This figure is obtained by adding the individual molar masses of <em><u>four oxygen atoms</u></em>, <em><u>one manganese atom</u></em> and <em><u>one potassium atom</u></em>

Explanation:

6 0
3 years ago
Read 2 more answers
Which air pollutant contributes to asthma?
Mnenie [13.5K]
The correct answer is the second option. Particulate matter is the air pollutant known to affect people with asthma. It also can cause other illnesses like cough, chest discomfort and a burning sensation in the lungs.
8 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 m
IrinaK [193]

Answer:

49.45~%

Explanation:

In this case, we have to start with the <u>chemical reaction</u>:

C_6H_1_2O~->~C_6H_1_0~+~H_2O

So, if we start with <u>10 mol of cyclohexanol</u> (C_6H_1_2O) we will obtain 10 mol of cyclohexanol (C_6H_1_0). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

(6*12)+(10*1)=82~g/mol

With this value we can calculate the grams:

10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0

Now, we have as a product 500 mL of C_6H_1_0. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

500~mL\frac{0.811~g}{1~mL}=405.5~g

Finally, with these values we can calculate the <u>yield</u>:

%~=~\frac{405.5}{820}x100~=~49.45%%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

6 0
3 years ago
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