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Elis [28]
3 years ago
9

The piece of iron that miguel measured had a mass of 51.1 g and a volume of 6.63 cm 3 . what did miguel calculate to be the dens

ity of iron?
Chemistry
1 answer:
likoan [24]3 years ago
4 0
Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³

By definition, 
Density = Mass/Volume
              = (51.1 g)/(6.63 cm³)
              = 7.7074 g/cm³

In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
              = 7707.4 kg/m³

Answer: 7.707 g/cm³ or 7707.4 kg/m³

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EXTRA CREDIT
marin [14]

Answer:

\huge\boxed{\sf 36\ H\ atoms}

Explanation:

<u>Molecular formula from Glucose:</u>

C₆H₁₂O₆

<u>3 moles of Glucose:</u>

3C₆H₁₂O₆

In 1 mole of Glucose, there are 12 hydrogen atoms.

<u>In 3 moles:</u>

= 12 × 3

= 36 H atoms

\rule[225]{225}{2}

5 0
2 years ago
Explain why is it better to use images on hazard labels, rather than words
enot [183]
In the case of an emergency where you might not have enough time to read several lines of writing, not to mention trying to find the hazard warnings when the whole bottle is probably covered in writing, it is much easier to locate and read universal hazard symbols.
5 0
3 years ago
Read 2 more answers
Which reaction has the highest atom economy for the production of C02?​
In-s [12.5K]

The highest atom economy

2CO + O₂ ⇒ 2CO₂

<h3>Further explanation</h3>

Given

The reaction for the production of CO₂

Required

The highest atom economy

Solution

In reactions, there are sometimes unwanted products that can be said to be a by-product or a waste product. Meanwhile, the desired product can be said to be a useful product, which can be shown as the atom economy

of the reaction

the higher the atomic economy value of a reaction, the smaller the waste/ byproducts produced, so that less energy is wasted

The general formula:

Atom economy = (mass of useful product : mass of all reactants/products) x 100

<em>or </em>

Atom economy = (total formula masses of useful product : total formula masses of all reactants/products) x 100

So a reaction that only produces one product will have the highest atomic value, namely the reaction in option C

4 0
2 years ago
3.75 g of an unknown gas at 59 °C and 1.00 atm is stored in a 1.35-L flask. What is the molar mass of the gas?
MAXImum [283]
You need to find moles of the gas, so you would use the ideal gas law:
PV=nRT
Pressure
Volume
n=moles
R= gas constant
Tenperature in Kelvin
n= PV/RT
(1.00atm)(1.35L)/(.08206)(332K) = 0.050mol
Molar mass is grams per mole, so
(3.75g/.050mol) = 75g/mol
4 0
3 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
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