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jonny [76]
3 years ago
12

Giovanni created this chart to study for an exam. Which best describes how to correct the error in the chart? Remove “plastics”

from the uses for sodium hydroxide. Swap the uses for iron oxide and calcium carbonate. Change the use for calcium chloride to “used in preservatives.” Add “de-ice roads” to uses for calcium carbonate.
Chemistry
2 answers:
Nastasia [14]3 years ago
6 0

Answer:

the answer is swap the uses for iron oxide and calcium carbonate. on edg

Vanyuwa [196]3 years ago
5 0

Calcium carbonate when mixed with water result in the formation of Calcium hydroxide which is used as an antacid. The chloride ion also helps in the whitening of teeth thus finding its use as an ingredient in toothpaste.

Iron oxide is dark pigment which is naturally used in makeup.

<span>Therefore the answer to this question is Swap the uses for iron oxide and calcium carbonate. </span>

<span> </span>

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What mass of carbon dioxide is formed when 1.75 mol of ethane burns completely in oxygen?
dybincka [34]

Answer:

There is a mass of 154 Grams of Carbon Dioxide.

Explanation:

One mole is equal to 6.02 × 10^23 particles.

This means we have 1.05 X 10^24 total particles of Ethane.

Each ethane particle contains 2 carbon atoms.

If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)

Carbon Dioxide has a molar mass of 44.01 g/mol

So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.

Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.

What you should end up with is 154 Grams of Carbon Dioxide.

Hope this helps (And more importantly I hope I didn't make any errors in my math lol)

As a side note this is all assuming that this takes place at STP conditions.

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3 years ago
Calculate the molarity of a solution if there are 1.5 mol of NaCl in 2.3 L of solution
frosja888 [35]

Answer:

0.6522 mol/L.

Explanation:

<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>

<em />

M = (no. of moles of solute)/(V of the solution (L)).

<em>∴ M = (no. of moles of solute)/(V of the solution (L)) </em>= (1.5 mol)/(2.3 L) = <em>0.6522 mol/L.</em>

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