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Romashka-Z-Leto [24]

2 years ago

8

In a test tube you add 2 mL of saturated chlorine water followed by about 1 mL of an organic solvent such as heptane (or cyclohe

xane). The test tube is shaken and you notice that the heptane layer has a yellow color. To this mixture you then add 1 mL of 0.1 M N a I. After shaking you notice a color change from yellow to purple in the heptane layer. What can you conclude about Cl2 and I minus from this experiment

1 answer:

bazaltina [42]2 years ago

7 0

Answer:

The organic compounds (heptane/cyclohexane) will get dissolve in Cl₂ water forming a yellow colored solution.

When NaI is added into it, the following reaction between NaI and Cl2 takes place;

Cl₂ + NaI ⇒ I₂ + NaCl

In above reaction, oxidation state of Cl is getting reduced from 0 in Cl2 to -1 in NaCl.

Therefore, Cl2 is getting reduced and acting as an oxidizing agent in the reaction. While I⁻ is getting oxidised annd acting as a reducing agent

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The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters

Ivenika [448]

Answer:- $4.15*10^-^2^0\mu L$

Solution:- It is a volume unit conversion problem where we are asked to convert the volume from $cm^3$ to microliters.

We know that:

$1cm^3$ = 1 mL

$1mL=10^-^3L$

and, $1L=10^6\mu L$

Let's use these conversions factors for the desired conversion using dimensional as:

$4.15*10^-^2^3cm^3(\frac{1mL}{1cm^3})(\frac{10^-^3L}{1mL})(\frac{10^6\mu L}{1L})$

= $4.15*10^-^2^0\mu L$

So, the answer is $4.15*10^-^2^0\mu L$ .

7 0

2 years 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

2 years ago

Which two rocks are primarily composed of a mineral that bubbles with acid?

Marianna [84]

To most geologists, the term "acid test" means placing a drop of dilute (5% to 10%) hydrochloric acid on a rock or mineral and watching for bubbles of carbon<span> dioxide gas to be released. The bubbles signal the presence of carbonate minerals such as</span>calcite<span>, </span>dolomite<span>, or one of the minerals listed in Table 1.</span>

8 0

2 years ago

Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2:

irina1246 [14]

Answer:

0.5

Explanation:

2NaCl(s) + 2H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2H2O(g) + Cl2(g)

Using ideal gas equation,

PV = nRT

28.7torr

Converting torr to atm,

= 0.0378atm

V = 0.597L

T = 27 °C

= 300 K

a) PV = nRT

(0.0378atm) * (0.597L) = n(0.0821) * (300k)

= 0.000915 mol

moles of water and chlorine = 0.000915 mol

From the above equation, the ratio of water to chlorine = 1 : 2

Therefore, mole of chlorine = 0.000915/2

= 0.000458 mol

mole fraction = moles of specie/moles of all the species present

= 0.000458/0.000915

= 0.5

5 0

2 years ago

At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?

Gnom [1K]

The temperature that would the volume of a gas be 0.550l if it had a volume of 0.432 L at -20.0 c is calculated using the Charles law formula

that is v1/T1=V2/T2
V1=0.550 l
t1=?
T2= -20 c +273 = 253 K
v2= 0.432 l

by making T1 the subject of the formula T1= V1T2/V2

T1= (0.55lL x253)/ 0.432 l = 322.11 K or 322.11-273 = 49.11 C

8 0

2 years ago

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