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3 years ago

How many grams of O2 are produced as 0.033 mol of water decompose

Chemistry

1 answer:

PolarNik [594]3 years ago

6 0

Answer: This can be quickly solved with "traintracks"

Explanation:

You start w/ grams of water and want to find moles of oxygen gas produced.

So you want to Convert:

Grams of water -> moles of water -> moles of oxygen gas.

The two things you need to know to set up the tracks are:

1)Molar mass of water- H2O

Hydrogen - 1.008(x2)

Oxygen - 16.00

Water - 18.016

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If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

Schach [20]

Answer: The amount of water required to prepare given amount of salt is 398.4 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

$0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL$

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

4 0

3 years ago

Hemoglobin is the protein that transports oxygen in mammals. Different species of mammals have slightly different forms of hemog

Anna35 [415]

Answer;

= 64561.95 g/mole

Explanation;

mass of Fe in 100g = .346g

= .346 / 55.8452 moles

= 0.0061957 moles

These represent 4 moles of Fe in the molecule so moles of hemaglobin

= 0.0061957/4

= 0.0015489 moles

these are in 100 g so mass of 1 mole = 100 / 0.0015489

= 64561.95 g / mole

molar mass of hemoglobin = 64561.95 g/mole

6 0

3 years ago

How many nonmetals are there on the periodic table?

ZanzabumX [31]

There are 17 nonmetals

4 0

3 years ago

Read 2 more answers

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

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

5 0

3 years ago

Atoms of the element beryllium would most likely behave similar to the way _______ behaves.

jasenka [17]

I would say a. lithium but i am not sure.

7 0

3 years ago