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kvasek [131]
2 years ago
6

The table shows columns that Brenda uses for her notes on the properties of elements. Her notes state that some elements can rea

ct to form basic compounds. Where should Brenda place this property in her table? only in the column for metalloids
Chemistry
2 answers:
skad [1K]2 years ago
8 0

Answer:

  • <u>in the columns for metals and for metalloids</u>

Explanation:

There are six elements that are always classified as metalloids: boron, silicon, germanium, arsenic, antimony, and tellurium. Pollonium is also, generally, classified as a metalloid,

Metalloids have intermediate electronegativity values (in between that of metals and nonmetals), which is responsible for some similarities (or in between properties) with metals and some similarites with non metals.

An example of such properties that metals and metalloids have in common is that they have relative high melting points. Metalloids are all solid at room temperature, such as most metals.

Other property that both metals and metalloids share is that they can react with oxygen to form oxides that are amphoteric.

Amphoteric compounds are substances that can behave as a base or as an acid, depending on the other compound with which they react.

For instance, among metal oxides, aluminum hydroxide, Al(OH)₃, will act as a base when reacts with hydrochloric acid, HCl, and will react as an acid when reacts with sodium hydroxide, NaOH.

The oxides of metalloids are usually amphoteric.

anzhelika [568]2 years ago
8 0

Answer:D

Explanation:

Took test

PLZ VOTE ME BRAINIEST

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Almost all corn oil is expeller-pressed

Explanation:

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20 point, pls help. A 5.50 mole sample of a gas has a volume of 2.50 L. What would the volume be if the amount increased to 11.0
alexandr1967 [171]

Answer:

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1 year ago
A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and
Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

0.03337 M=\frac{n}{0.02328 L}

n = 0.03337 M\times 0.02328 L=0.0007768 mol

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

\frac{1}{1}\times 0.0007768 mol=0.0007768 mol chloride ions.

In one mole of aldrin there are 6  moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

\frac{0.0007768 mol}{6}=0.0001295 mol

Moles of aldrin present in the sample = 0.0001295 mol

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8 0
2 years ago
A chemist is working on a reaction represented by this chemical equation:
erastova [34]

Answer is: A. 1.81 mol.

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n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.

n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.

From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

n(Fe(OH)₂) = 3.62 mol ÷ 2.

n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.

6 0
2 years ago
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Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod
son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0
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