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3 years ago

7

Why parafin oil use in melting

1 answer:

MatroZZZ [7]3 years ago

4 0

Answer:

Paraffin oil is used for determination of boiling point and melting point for the following reasons: It has a very high boiling point and so it can be used to maintain high temperatures in the boiling and melting point apparatus without loss of the substance.

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HELP ME PLEASE. no links.

tatiyna

Answer:

Experimental group

Explanation:

i hopes this helps let me know if im wrong:)

4 0

3 years ago

You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How

Ksju [112]

Answer:

Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.

Explanation:

To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask

The molar mass of Aluminum Sulfate = 342.15 g/mol

Using the molarity formula:-

Molarity = Number of moles/Volume of solution in a liter

Number of moles = Given weight/ molar mass

Molarity = (Given weight/ molar mass)/Volume of solution in liter

0.187 M = (Given weight/342.15 g/mol)/0.250 L

Given weight = 15.99 g

8 0

3 years ago

A nitrogen atom in N2 should have a

SIZIF [17.4K]

Answer:

The answer is partial charge 8+.

7 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

____________ enables you to ride a bike without skidding and falling.

Marina86 [1]

Sliding or rolling friction

4 0

3 years ago