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3 years ago

Give the reason why the empty crucible should be heated before starting the experiment

1 answer:

3 0

To make sure no cracks are in the crucible and also to remove any moisture present in the crucible by the process of heating.

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Students investigate forces and how they affect objects. They use a small car and wooden blocks to carry out their investigation

Umnica [9.8K]

Answer:

increase the number of blocks

increase the number of students pulling the cart

increase the number of blocks and increase the number of students pulling the cart

decrease the number of blocks

decrease the number of students pulling the cart

decrease the number of blocks and decrease the number of students pulling the cart

(ANSWER CHOICES)

Explanation:

7 0

3 years ago

Determine the formal charge on each atom in the structure.

jok3333 [9.3K]

Answer:

−0.75 , 1.25

Explanation:

Please refer to attachment for more information

7 0

3 years ago

Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but

alukav5142 [94]

Answer:

a) For nicotine, the protonated form is the present in stomach.

b) For caffeine, the neutral base form is the present in stomach.

c) For Strychinene, the protonated form is the present in stomach.

d) For quinine, the protonated form is the present in stomach.

Explanation:

In a basic dissociation for molecules with basic nitrogen, the equilibrium is:

A + H₂O ⇄ AH⁺ + OH⁻

<em>Where A is neutral base and AH⁺ is protonated form</em>

The basic dissociation constant, kb, is:

$K_{b} = \frac{[AH^+][OH^-]}{[A]}$

As pH in stomach is 2,5:

[OH] = $10^{-[14-pH]}$

[OH] = 3,16x10⁻¹² M

Thus:

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$ <em>(1)</em>

Using (1) it is possible to know if you have the neutral base or the protonated form, thus:

(a) nicotine Kb = 7x10^-7

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$221359 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For nicotine, the protonated form is the present in stomach

(b) caffeine,Kb= 4x10^-14

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$0,0127 = \frac{[AH^+]}{[A]}$

[AH⁺}<<<<[A]

For caffeine, the neutral base form is the present in stomach

(c) strychnine Kb= 1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$314456 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For Strychinene, the protonated form is the present in stomach

(d) quinine, Kb= 1.1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$347851= \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For quinine, the protonated form is the present in stomach.

I hope it helps!

7 0

3 years ago

For the reaction below determine the ▲H for the reaction and state whether the reaction was endothermic or exothermic. Show your

scoundrel [369]

Answer:exothermice

Explanation: it is relaeing heat not keeping it in

6 0

2 years ago

A sample of an unknown liquid has a volume of 30.0 mL and a mass of 6 g. What is its density? please Show your work or explain h

Leno4ka [110]

The density of the liquid is 0.2 g/mL.

The mass of the liquid is 6 g.

The volume of the liquid is 30.0 mL.

Density = mass/volume = 6 g/30.0 mL = 0.2 g/mL

3 0

3 years ago