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Luba_88 [7]
2 years ago
9

In the compound aluminum oxide, which is the cation?

Chemistry
1 answer:
8090 [49]2 years ago
7 0

Answer:

-Aluminum Oxide: The cation is Al3+ and the anion is O2-. The sum of the charges for aluminum oxide is 2(3+) + 3(2-) = 0. Thus, the formula is Al2O3. An ionic compound is named using the name of the cation followed by the name of the anion, eliminating the word ion from each.

Explanation:

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It is ionic bond because of Na valency and it is alkali metal and cl is non metal and it’s valency , they both share electron and make ionic bonding .
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Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m
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If 50.0 grams of HCl were dissolved in enough water to make a solution with a total volume of 5.0 liters, what would be the mola
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The molarity of a solution is found to be 10 M.

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                                        Litres solution

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Based on the enthalpy changes given for the reactions, which reactions are endothermic?
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What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0
2 years ago
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