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3 years ago

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I NEED HELP FAST !The density of aluminum is 2.70 g/cm3. How large a cube, in cm3, would contain 2.00 x 10^24 atoms of aluminum?

Use dimensional analysis to solve and show all work including units on every number!

1 answer:

IceJOKER [234]3 years ago

4 0

Explanation:

2.70 times 2.00 times 10...24=

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For the following amino acids, draw a triangle around the amine group, draw a square around the carboxyl and circle the side gro

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hooc are carboxyl groups

your r or amino groups are those unique structures which have different atoms in them. your nh2 groups are your hydrogen atoms

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3 years ago

For the following reaction,

jeka94

Answer:

The answer to your question is:

1.- CO

2.- 0.414 moles of CO2

Explanation:

Data

2CO + O2 ⇒ 2CO2

CO = 0.414 moles

O2 = 0.418

Process

theoretical ratio CO/O2 = 2/1 = 1

experimental ratio CO/O2 = 0.414/0.418 = 0.99

Then the limiting reactant is CO

2.-

2 moles of CO --------------- 2 moles of CO2

0.414 moles of CO --------- x

x = (0.414 x 2) / 2

x = 0.414 moles of CO2

5 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

4 years ago

A lake has been infected by some type of new algae that is unknown. Every single day the amount of surface area that the algae t

FrozenT [24]

Answer:

It takes 86 days take to cover half of the lake

Explanation:

In the day #1, the amount of the algae is X,

In the day #2 is 2X

In the day #3 is 2*2*X = X*2²

...

In the day #n the amount of the algae is X*2^(n-1)

Assuming X = 1m³. In the day 87, the area infected was:

1m³*2^(87-1)

7.74x10²⁵m³ is the total area of the lake

the half of this amount is 3.87x10²⁵m³

The time transcurred is:

3.87x10²⁵m³ = 1m³*2^(n-1)

Multiplying for 5 in each side:

ln (3.87x10²⁵) = ln (2^(n-1))

58.9175 = n-1 * 0.6931

85 = n-1

86 = n

<h3>It takes 86 days take to cover half of the lake</h3>

4 0

3 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago