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Luda [366]
2 years ago
8

A substance is 89.2% carbon by mass. how much of the substance would be needed to recover 34.6 mol of pure carbon?

Chemistry
2 answers:
kari74 [83]2 years ago
8 0
1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon, 
then, (415.2/89.2) ×100 =  465.47 g of the substance will be required to yield 34.6 moles of carbon.
daser333 [38]2 years ago
7 0

Answer:

The amount of substance we need to recover 34.6 mol of carbon is 465.47g

Explanation:

we want to know what mass of substance we need to recover 34.6 mol of carbon

This substance is composed of 89.2% carbon

Wmust recover 34.6 mol of carbon and we know that the molecular mass of carbon is 12g / mol this means that one mole of carbon has a mass of 12 g

We use a simple rule of three to know how many grams of carbon are in 34.6 mol

1 mol C\longrightarrow 12g\\34.6 mol C\longrightarrow x\\x=\frac{(34.6)(12)}{1}= 415.2 g

415.2 g are 34.6 mol of carbon

As the amount of carbon we want to recover is 415.2g (34.6mol)  and this amount corresponds to 89.2% of the total

To calculate the total mass we need we use a simple rule of three

89.2\%\longrightarrow 415.2 g\\100\%\longrightarrow x\\x=\frac{(100)(415.2)}{89.2}\\ x= 465.4g

The amount of substance we need to recover 34.6 mol of carbon is 465.47g

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Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis
almond37 [142]

Answer: a) BaCl_2

b)  BiO_3H_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = \frac{0.48}{0.48}=1

For O =\frac{0.96}{0.48}=2

The ratio of Ba: Cl= 1:2

Hence the empirical formula is BaCl_2

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles

Moles of O= \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= \frac{0.38}{0.38}=1

For O =\frac{1.15}{0.38}=3

For H=\frac{1.16}{0.38}=3

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is BiO_3H_3

6 0
2 years ago
A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

             = 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0
2 years ago
Explain why groups 14 and 15 are better representatives of mixed groups than groups 13 and 16
Brrunno [24]
<h3><u>Answer</u>;</h3>

Groups 14 and 15 each contain metals, nonmetals, and metalloids while Group 13 contains metals and a metalloid, and Group 16 contains metalloids and nonmetals.

<h3><u>Explanation;</u></h3>
  • Groups 13–16 of the periodic table contain one or more metalloids, in addition to metals, nonmetals, or both.
  • Unlike other groups of the periodic table, which contain elements in one class, groups 13–16 referred to as mixed groups contain elements in at least two different classes. In addition to metalloids, they also contain metals, nonmetals, or both.
  • <em><u>Group 14 also known as the carbon group contains carbon which is a non metal, silicon and germanium which are metalloids and tin and lead which are metals.</u></em>
  • <em><u>Group 15 also known as the Nitrogen group contains non metals such as oxygen, metalloid tellurium and a metal polonium.</u></em>
3 0
2 years ago
Elements that do not have full outer electron shells will donate, share, or take electrons from other atoms. Choose the items th
Tamiku [17]

it´s actually Lithium and fluorine / Magnesium and Chlorine / Beryllium and Nitrogen

8 0
2 years ago
Bismuth(III) sulfate is an ionic compound formed from Bi3+ and SO42-. What is the correct way to represent the formula?
Hitman42 [59]

Answer:

Bi2(SO4)3

Explanation:

Bismuth(iii) sulfate is an ionic compound therefore, their is transfer of electron. Ionic compound has both cations and anions. The cations is positively charged ion while the anions is negatively charged ions. The cations loses electron to become positively charged while the anions gains electron to become negatively charged.

From the compound above, Bismuth(iii) sulfate the cations will be Bismuth ion which loses 3 electrons. The anions is the sulfate ion (S04)2- with a -2 charge.

The chemical formula can be computed from the charge configuration as follows

Bi3+  and (SO4)2-

cross multiply the charges living the sign behind to get the chemical formula

Bi2(SO4)3

Note the final chemical formula, the numbers are sub scripted

4 0
2 years ago
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