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patriot [66]
2 years ago
10

How many g i2 should be added to 750g ccl4 to prepare a 0.200m solution?

Chemistry
2 answers:
vesna_86 [32]2 years ago
6 0
<span>There are a number of ways to express concentration of a solution. This includes molality. Molality is expressed as the number of moles of solute per mass of the solvent. We calculate as follows:

0.200 mol I2 / kg CCl4 ( .750 kg CCl4 ) ( 253.809 g I2 / mol I2) = 38.07 g I2 needed

Hope this helps.

 </span>
tamaranim1 [39]2 years ago
3 0

Answer:

Mass of I2 required = 38.1 g

Explanation:

<u>Given:</u>

Mass of CCl4 = 750 g = 0.750 kg

Molality of the solution = 0.200 m

<u>To determine:</u>

Mass of I2

<u>Formula:</u>

Molality of a solution is expressed as the ratio of the amount of solute per kg of solvent

Molality = \frac{moles\ of\ solute}{kg\ solvent} \\

Here the solute is I2 and solvent is CCl4

Moles\ of\ I2 = molality*mass\ solvent = 0.200*0.750= 0.150 moles\\\\Molar\  mass\  of\  I2 = 253.8 g/mol\\\\Mass\ of\ I2 = moles*molar\ mass = 0.150*253.8 = 38.1 g

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Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of
gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y <em>(1)</em>

<em>Where X is molar fraction of Al and Y is molar fraction of Li.</em>

X + Y = 1 <em>(2)</em>

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= <em>3,47%</em>

6 0
1 year ago
BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: K_{b} = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: K_{w} = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
1 year ago
If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?
solmaris [256]

Answer:

6.7 x 10²⁶molecules

Explanation:

Given parameters

Mass of CO₂  = 4.9kg  = 4900g

Unknown:

Number of molecules  = ?

Solution:

To find the number of molecules, we need to find the number of moles first.

 Number of moles  = \frac{mass}{molar mass}

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   Number of moles  = \frac{4900}{44}  = 111.36mole

A mole of substance is the quantity of substance that contains the avogadro's number of particles.

       1 mole  = 6.02 x 10²³molecules

     111.36 moles  =   111.36 x 6.02 x 10²³molecules   = 6.7 x 10²⁶molecules

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bearhunter [10]
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Kp = [PCl₅]/[PCl₃][Cl₂]

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2 years ago
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