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3 years ago

Can someone help me with this

Chemistry

2 answers:

topjm [15]3 years ago

8 0

Answer: It would be malleable, solids, luster, conductors, reactive

Explanation:

Sedbober [7]3 years ago

5 0

Answer:

Explanation:

Please give me brainliest :)

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using the following thermochemical equation, determine the amount of heat produced from the combustion of 24.3 g benzene (c6h6).

aliya0001 [1]

the amount of heat produced from the combustion of 24.3 g benzene (c6h6) is ΔH = -976.5 kJ

There are two moles of benzene involved in the process (C6H6). Since the heat of this reaction is -6278 kJ, the burning of 2 moles of benzene will result in a heat loss of 6278 kJ. This reaction is exothermic.

Enthalpy, or the value of H, is a unit of measurement for heat that relies on the amount of matter present (number of moles).

Thus, 24.3 g of benzene contains:

n = mass/molar mass, where n = 24.3/78.11, and n = 0.311 moles.

2 moles = 6278 kJ

0.311 moles =x

By the straightforward direct three rule:

2x = -1953.08 x = -976.5 kJ

Learn more about combustion here-

brainly.com/question/15117038

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4 0

1 year ago

Provide the name(s) for the tertiary alcohol(s) with the chemical formula C6H14O that have a 4-carbon chain. Although stereochem

pochemuha

Answer:

Explanation:

A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.

Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is

2-hydroxy-2,3-dimethylbutane

H OH H H

| | | |

H - C - C - C - C - H

| | | |

H CH₃ CH₃ H

From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain

4 0

3 years ago

I would appreciate your help! Due today! ( I don't know if this is even chemistry)

Vinil7 [7]

Answer:

6+69

Explanation:

6 0

3 years ago

Read 2 more answers

How many moles are in 25.2g of KMnO4?

Zolol [24]

When finding the moles in a compound you have to know the grams. In this case, 25.2 grams are given for KMnO4. To find the moles you would divide the amount of grams by the molar mass of KMnO4. The molar mass of KmnO4 is 158.034. You you would now divide 25.2 by 158.034 which is 0.15946 moles. Depending on how many decimal places the questions asks for is dependent on you. I just went with 5 significant figures.

4 0

3 years ago

Read 2 more answers

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$