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2 years ago

10. If 3.5 kJ of energy are added to a 28.2 g sample of iron at 20°C, what

Chemistry

1 answer:

igor_vitrenko [27]2 years ago

3 0

Answer:

569K

Explanation:

Q = 3.5kJ = 3500J

mass = 28.2g

∅1 = 20°C = 20 + 273 = 293K

∅2 = x

c = 0.449

Q = mc∆∅

3500 = 28.2×0.449×∆∅

3500 = 12.6618×∆∅

∆∅ = 3500/12.6618

∆∅ = 276.4220

∅2 - ∅1 = 276.4220

∅2 = 276.4220 + ∅1

∅2 = 276.4220 + 293

∅2 = 569.4220K

∅2 = 569K

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Why did mendeleev leave blank spots in his periodic table?

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He thought elements that haven't been discovered belonged in the place of the gap. He could also use the atomic mass of the missing elements

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2 years ago

Which is something that a PURE SUBSTANCE and a SOLUTION have in common?

zhuklara [117]

D) They both look uniform (the same) throughout.

<h3>Further explanation</h3>

Pure substance can be any element or compound and is formed from one type of atom/molecule only

Meanwhile, the solution is included in a mixture consisting of 2 or more pure substance

Pure substance can be formed through a chemical process while the mixture is through a physical process

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2 years ago

4. Synthesis problems. Propose a synthetic route from the starting material to the product. In your answer, show each individual

Troyanec [42]

The reduction of alkyne to an alkene in the first step allows the best reagent to be chosen for each subsequent step.

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1 year ago

The density of copper is 8.92 g/mL. The mass of a piece of copper that has a volume of 10.3 mL is

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Answer:

The answer is

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of copper = 10.3 mL

density = 8.92 g/mL

The mass is

mass = 8.92 × 10.3 = 91.876

We have the final answer as

Hope this helps you

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3 years ago

For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.

Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

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2 years ago