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2 years ago

How many particles are present in0.24moles of carbon?

Chemistry

1 answer:

Ray Of Light [21]2 years ago

7 0

Answer:

1.45 x 10²³ particles

Explanation:

Given parameters:

Number of moles of carbon = 0.24moles

Unknown:

Number of particles = ?

Solution:

A mole of a substance contains the Avogadro's number of particles.

The Avogadro's number of particles is 6.02 x 10²³

So;

0.24 moles of carbon will contain 0.24 x 6.02 x 10²³ = 1.45 x 10²³ particles

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Which reactant is unlikely to produce the indicated product upon strong heating?

Ilya [14]

2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.

Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.

A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.

Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.

Full question :

Q. Which reactant is unlikely to produce the indicated product upon strong heating?

A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
B) 2-Ethylpropanedioic acid Butanoic acid
C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone

Hence, option (D) is correct.

Learn more about carboxylic acid here : brainly.com/question/26855500

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8 0

1 year ago

Using off-road vehicles does not contribute to the process of erosion. true or false

Artemon [7]

Using off road vehicles does help contribute to the process of erosion.

7 0

2 years ago

Read 2 more answers

You have used 2.4×102 L of distilled water for a dialysis patient. How many gallons of water is that?

Alinara [238K]

Answer:

63.36gallons

Explanation:

Given:

Volume of water used for dialysis = 2.4 x 10²L

Solution:

We are to convert from liters to gallons.

The conversion factor is shown below:

1L = 0.264gallons

To convert to litre:

since 1L = 0.264gallons

2.4 x 10²L = (2.4 x 10² x 0.264)gallons; 63.36gallons

3 0

3 years ago

Significant digits are the number of digits that reflect the precision of a measurement or number.

Delvig [45]

Answer:

True

Explanation:

Significant digits are numbers that helps to present the precision of measurements calculations.

Numbers that do not contribute to the precision of a reading should not be counted as significant.

There are rules of assigning significant numbers:

Leading or trailing zeros are insignificant and should only be counted as a place holder.
All non-zero digits are significant
Zeroes between non-zero digits are significant.
Leading zeros in a decimal are significant before the number.
All the numbers in a scientific notation are significant.

5 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago