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3 years ago

3. How many moles are in 1.49 x 1023 molecules of iodine?

Chemistry

1 answer:

tino4ka555 [31]3 years ago

7 0

Answer:

<h3>The answer is 0.25 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{1.49 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{1.49}{6.02} \\ = 0.247508305...$

We have the final answer as

<h3>0.25 moles</h3>

Hope this helps you

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According to the Kinetic Molecular Theory, all substances contain entities that are in a continuous random motion.Could you plea

Romashka-Z-Leto [24]

The answer is a bit ambiguous since the kinetic molecular theory refers to the behavior of gas molecules, and says that these are in continuous and random motion. All substances with certain conditions of temperature and pressure can be taken to a gaseous state, so I could tell you that this answer is true.

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1 year ago

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

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4 years ago

Type the correct answer in the box. Express your answer to two significant figures. You are performing a reaction with 1.7 moles

konstantin123 [22]

Answer:

1.34 34343 he did

Explanation:

i did the test

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3 years ago

Naturally occurring copper has two isotopes, 63cu and 65cu. what is different between atoms of these two isotopes?

GaryK [48]

The differences are <u>the number of neutrons</u> and the <u>atomic mass</u><u /><u />.

Copper-63 has an atomic mass of 63 amu, and has 34 neutrons.
Copper-65 has an atomic mass of 65 amu, and has 36 neutrons.

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3 years ago

2. What is the name of the compound AIPO?

Andru [333]

Answer:

Aluminum phosphate

Explanation:

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3 years ago