ELEMENTS ARE CLASSIFIED AS METALS, NONMETALS, AND METALLOIDS. USE THE PERIODIC TABLE AND IDENTIFY TWO OF EACH CLASSIFICATION.

Chemistry

1 answer:

alekssr [168]2 years ago

3 0

Answer:

Elements are classified into- metals – Nonmetals- Metalloids – Noble gases. State which of A, B, C, D is a:

1) Metallic element

2) Non-metallic element

3) Metalloid

4) Noble gas.

A) Is non-malleable, non-ductile and a poor conductor of electricity

B) Has lustre, is malleable and ductile and a good conductor of electricity

C) Is unreactive and inert and present in traces in air

D) Shows properties of both metals and nonmetals

Explanation:

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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 %.

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1 year ago

One quarter of a 1-mole piece of iron turns to rust in six months. Which rate describes the speed at which the iron rusts?

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1/4 mol = 0.25 mol

6 months = 0.5 year

rate = 0,25 mol / 0.5 year = 0.5 mol/year or approx 0.042 mol/month

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