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Kazeer [188]
3 years ago
13

Name the unit being abbreviated in each measurement (spelling counts). Only write the name of the unit, do not include the numbe

r. 0.8 L litre 3.6 cm centimeter 4.0 kg kilogram 3.5 s 373 K
Chemistry
1 answer:
givi [52]3 years ago
5 0

Answer:

Litre (L) , Centimetre (cm) , Kilogram (Kg), Seconds (s) and Kelvin (K)

Explanation:

The units are used for the following measurement;

Litre = Volume

Centimetre = Length

Kilogram = Mass

Seconds = Time

Kelvin = Temperature

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Which produces the greatest number of ions when one mole dissolves in water?
Arada [10]
~Hello there!

Your question: Which produces the greatest number of ions when one mole dissolves in water?

Your answer: Na2SO4<span> can dissociate into three ions, whereas the other choices produce only two ions (or in the case of sucrose, none) when dissolved in water.

Any queries ^?

Happy Studying!
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7 0
3 years ago
Acetylene (c2h2) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[co2(g)] = –393.5
Brums [2.3K]
Answer is: 13181,7 kJ of energy <span>is released when 10.5 moles of acetylene is burned.
</span>Balanced chemical reaction: C₂H₂ + 5/2O₂ → 2CO₂ + H₂O.
<span>ΔHrxn = sum of ΔHf (products of reaction) - sum of ΔHf (reactants).</span><span>
Or ΔHrxn = ∑ΔHf (products of reaction) - ∑ΔHf (reactants).
ΔHrxn - enthalpy change of chemical reaction.
<span>ΔHf - enthalpy of formation of reactants or products.
</span></span>ΔHrxn = (2·(-393,5) + (-241,8)) - 226,6 · kJ/mol.
ΔHrxn = -1255,4 kJ/mol.
Make proportion: 1 mol (C₂H₂) : -1255,4 kJ = 10,5 mol(C₂H₂) : Q.
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8 0
3 years ago
Nitrogen and hydrogen react to produce ammonia. What mass of ammonia could be produced from 500 grams of nitrogen? Assume that e
OlgaM077 [116]
The answer is 607g. the working is shown above.

5 0
3 years ago
A student balances the following redox reaction using half-reactions.
ICE Princess25 [194]
Answer: 6.

Explanation:

1) Aluminum

Al^0-3e^----\ \textgreater \ Al^{3+}

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.

2) Manganesium

Mn^{2+}+2e^{-}---\ \textgreater \ Mn

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}&#10;&#10;3Mn^{2+}+6e^{-}---\ \textgreater \ 3Mn^{0}

4) Net equation

Add the two half-equations:

2Al^{0}+3Mn^{2+}----\ \textgreater \ 2Al^{3+}+3Mn^{0}

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.

The right side has 2 Al, 3 Mn, and 2*3 positive charges.

So, the equation is balanced.

5) Count the number of electrons involved.

As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.
5 0
3 years ago
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You would have spent $18.9
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3 years ago
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