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kolbaska11 [484]

3 years ago

8

The first step in the two-step process of making a solution is the breakdown of the solute source into ____________."

1 answer:

Troyanec [42]3 years ago

7 0

<h3><u>Answer;</u></h3>

Atomic particles

The first step in the two-step process of making a solution is the breakdown of the solute source into <u>atomic particles</u>.

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

Solution is a homogeneous mixture of two or more substances uniformly dispersed throughout a single phase
<em><u>The liquid component that makes up the largest part of a solution is called the solvent, and the substance that is dissolved is called the solute.</u></em>
Unsaturated Solution is a solution that contains less solute than a saturated solution does and that is able to dissolve additional solute, while saturated Solution is a solution that cannot dissolve any more solute under the given conditions

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pogonyaev

1/6 + 1/8 = 7/24

24

≅ 0.2916667

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3 years ago

How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -><br> 4A1 + 302

Evgesh-ka [11]

<h3>Answer:</h3>

$\displaystyle 40 \ mol \ Al$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 40 \ mol \ Al$

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0

2 years ago

Enter the correct 4 digit code (no spaces)

Airida [17]

3618 is the numbers for the graph

3 0

3 years ago

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

3 years ago

Read 2 more answers

A 50.0-g sample of liquid water at 25.0 Â°c is mixed with 23.0 g of water at 79.0 Â°c. the final temperature of the water is ___

I am Lyosha [343]

<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>

<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>

<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>

<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>

<span>73.0<span>Tf</span>=2561 °C</span>

<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>

8 0

3 years ago