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3 years ago

All circuits must have three things: energy source, wire, and a generator Lesson 5.04 Question 7 options: True False

Chemistry

1 answer:

Doss [256]3 years ago

5 0

Answer:

True

Explanation:

Sorry i'm late probably but i answered this for you :) i hope this helps though

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Determine the oxidation number of the element "J" in H3JO2-

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Answer:

Oxidation number:

3*1+ oxidation number of J+2*-2= -1

Oxidation number of J = 0

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2 years ago

What is a common form for synthesis reactions?

mamaluj [8]

Answer:

The Best Answer would be B but the correct answer is A+B --> AB

Explanation:

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2 years ago

What volume of O2 is needed to combust 425 mL of butane?

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Answer:

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3 years ago

Which of the following is an inexhaustible energy resource?

koban [17]

<u />C. Water is an inexhaustible energy resource among these options. Coal, oil, and natural gas we can run of, but for the foreseeable future, there will always be water.

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2 years ago

If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of

Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.1\ g}{186.04\ g/mol}$

$Moles\ of\ ferrocene= 0.0005375\ mol$

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

$\rho=\frac{Mass}{Volume}$

or,

$Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g$

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.044\ g}{78.49\ g/mol}$

$Moles\ of\ acetyl\ chloride= 0.0005606\ mol$

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms 0.0005375 mole of monoacetylferrocene

Moles of product formed = 0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0

3 years ago