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3 years ago

What is each vertical column of elements in the periodic table called?

1 answer:

7 0

The vertical (up and down) is the groups on the periodic table. Elements in a group shares similar properties.

Answer: Group

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An aluminum ion has a +3 charge. What change did the atom undergo to attain this charge?

lesya692 [45]

When an ion loses electrons, it becomes a positive ion even though it does not lose protons. If there are more protons than electrons in the ion, the ion will have a positive charge.

8 0

3 years ago

0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2Al(OH)3(s) +

Veseljchak [2.6K]

Answer:

The answer to your question is 1.1 moles of water

Explanation:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

0.45 mol 0.55 mol ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

3 moles of H₂SO₄ ---------------- 6 moles of water

0.55 moles of H₂SO₄ ----------- x

x = (0.55 x 6) / 3

x = 3.3 / 3

x = 1.1 moles of water

5 0

3 years ago

For which of the following processes would you expect there to be an increase in entropy? Ag+(aq) + Cl-(aq) AgCl(s) H2O(g) H2O(l

Lyrx [107]

CaBr₂(s) → Ca⁺²(aq) + 2Br⁻ (aq) ΔS>0

6 0

2 years ago

Read 2 more answers

True or False: Both molecules and compounds are pure substances.

Solnce55 [7]

Answer:

False

Explanation:

Hope this helps!

4 0

2 years ago

Read 2 more answers

What is the % composition of Carbon in Chromium (iii) Carbonate

photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

$Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3$

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

$\begin{gathered} C\rightarrow3\times12=36 \\ \\ O\rightarrow9\times16=144 \\ \\ Cr\rightarrow2\times52=104 \end{gathered}$

The molar mass will be thus:

$M=36+104+144=284\text{ g/mol}$

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

$\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\ \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}$

The percent composition of Carbon is thus 12.7 %.

8 0

1 year ago