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Gelneren [198K]
4 years ago
6

Chalcopyrite is an ore with the composition cufes2. what is the percentage of iron in a 39.6 g sample of this ore? answer in uni

ts of %.
Chemistry
1 answer:
iren2701 [21]4 years ago
5 0
<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass. For this, we look at the atomic masses of the elements present in the compound. Cu has an atomic mass of 63.546 amu Fe has 55.845 amu and S has 36.065 amu Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu So we have not established the mass of the compound in amus 63.546 + 55.845 + 72.13 = 191.521 That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845 So to get the percentage, or fraction of iron, we take 55.845 / 191.521 Which comes out to 29.15% by mass Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
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A wave has a frequency of 35 Hz and a wavelength of 15 meters, what is the speed of the wave?
puteri [66]

Answer:

1174.392 mph

Explanation:

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7 0
3 years ago
When 38.1 grams of a certain metal at a temperature of 90.0°C is added to 100.0mL of water at a temperature of 17.6°C in a perfe
MakcuM [25]

Answer:

a. qm = 627.3 J

b. qw = 627.3 J

c. C₂ = 227.4 J/kg.°C

Explanation:

a.

Since, the calorimeter is completely insulated. Therefore,

Heat Lost by Metal = Heat Gained by water

qm = qw

qm = m₁C₁ΔT₁

where,

qm = heat lost by metal = ?

m₁ = mass of water = (density)(volume) = (1000 kg/m³)(100 mL)(10⁻⁶ m³/1 mL)

m₁ = 0.1 kg

C₁ = specific heat capacity of water = 4182 J/kg.°C

ΔT₁ = Change in Temperature of Water = 19.1°C - 17.6°C = 1.5°C

Therefore,

qm = (0.1 kg)(4182 J/kg.°C)(1.5°C)

<u>qm = 627.3 J</u>

<u></u>

b.

Since,

qm = qw

<u>qw = 627.3 J</u>

<u></u>

c.

qm = m₂C₂ΔT₂

where,

m₂ = mass of metal = 38.1 g = 0.0381 kg

C₂ = specific heat capacity of metal = ?

ΔT₂ = Change in Temperature of metal = 90°C - 17.6°C = 72.4°C

Therefore,

627.3 J = (0.0381 kg)(C₂)(72.4°C)

(627.3 J)/(0.0381 kg)(72.4°C) = C₂

<u>C₂ = 227.4 J/kg.°C</u>

6 0
3 years ago
A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.
Alenkinab [10]

Answer:

\Delta V = 234.736\,mL

Explanation:

The quantity of moles of ethanol in the solution is:

n_{C_{2}H_{5}OH} = \left(\frac{597\,mL}{1000\,mL} \right)\cdot \left(8.35\,\frac{mol}{L} \right)

n_{C_{2}H_{5}OH} = 4.985\,mol

The mass and volume of ethanol in the solution are, respectively:

m_{C_{2}H_{5}OH} = (4.985\,mol)\cdot \left(46.07\,\frac{g}{mol} \right)

m_{C_{2}H_{5}OH} = 229.658\,g

V_{C_{2}H_{5}OH} = \frac{229.658\,g}{0.7893\,\frac{g}{mL} }

V_{C_{2}H_{5}OH} = 290.964\,mL

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

\Delta V = (525\,mL+597\,mL) - (597\,mL + 290.964\,mL)

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8 0
3 years ago
What action would you expect an atom of Aluminum to undergo?
KATRIN_1 [288]

Answer:

C. Lose three electrons to have a full outer shell  

Explanation:

Al is in Group 13 of the Periodic Table, so it has three valence electrons.

It must either lose three electrons or gain five to achieve a stable octet.

It is easier to lose three electrons than it is to gain five, so Al loses three electrons.

D. is wrong, for the same reason.

A. is wrong. If Al lost three electrons, it would be breaking into a stable inner shell.

C. is wrong. Al is a metal, so it will lose electrons in a reaction.

6 0
4 years ago
What is the limiting reactant when 19.9 g CuO react with 2.02 g H2?
Harlamova29_29 [7]

Answer:

Explanation:

use the equation

moles = mass/mr

=19.9/79.5

=0.250moles of CuO

then do the same for

H = 2.02/1

=2.02

so CuO is the limiting reagent because there is less amount of it.

Hope this helps  :)

4 0
3 years ago
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