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2 years ago

The element in group 4 and period 3

2 answers:

8 0

The three group 4 elements that occur naturally are titanium, zirconium, and hafnium. The first three members of the group share similar properties; all three are hard refractory metals under standard conditions.

olganol [36]2 years ago

7 0

the answer is

Silicon

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How many atoms are there in 5.00 mol of sulphur (S)?

Mumz [18]

Answer:

3.01 × 10²⁴ atoms S

General Formulas and Concepts:

Chemistry - Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

5.00 mol S

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$5.00 \ mol \ S(\frac{6.022 \cdot 10^{23} \ atoms \ S}{1 \ mol \ S} )$ = 3.011 × 10²⁴ atoms S

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

3.011 × 10²⁴ atoms S ≈ 3.01 × 10²⁴ atoms S

6 0

3 years ago

A crate holds 20 boxes of sewing supplies. Each box contains 8 spools of thread. Each spool has 250 meters of thread wrapped aro

exis [7]

Answer: 40000

Explanation:

250*8=2,000

2000*20=40000

5 0

1 year ago

Which of the following are likely to form a covalent bond?

Marat540 [252]

hydrogen and oxygen atoms

Explanation:

this is because they are non metals and there will be sharing of electrons between the two atoms forming the bond

4 0

3 years ago

Which of the following metals may tarnish to a greenish-blue color called verdigris

hoa [83]

Copper. thats the way copper oxidized. like the statue of liberty

3 0

3 years ago

Read 2 more answers

Place the following transitions of the hydrogen atom in order from longest to shortest wavelength of the photon emitted. Rank fr

gizmo_the_mogwai [7]

Answer:

7 to 4 (higher $\lambda$ )

5 to 3

3 to 2

4 to 2 (lower $\lambda$ )

Explanation:

We can use the Rydberg formula which relates the wavelength of the photon emissions to the principle quantum numbers involved in the transition:

$\frac{1}{\lambda}=R((\frac{1}{n_1^2})-(\frac{1}{n_2^2}))$

with $n_1$ final n, and $n_2$ initial n

evaluating for each transition:

7 to 4 $\frac{1}{\lambda} = R((\frac{1}{4^2})-(\frac{1}{7^2}))= R(1/16-1/49)=R(0.042)$

5 to 3 $\frac{1}{\lambda}= R(1/9-1/25)=R(0.071)$

4 to 2 $\frac{1}{\lambda}= R(1/4-1/16)=R(0.1875)$

3 to 2 $\frac{1}{\lambda}= R(1/4-1/9)=R(0.139)$

Note that the above formula is written for $\frac{1}{\lambda}$ , so lower $\frac{1}{\lambda}$ value obtained involves higher $\lambda$ .

So we should order from lower to higher $\frac{1}{\lambda}$

7 to 4 (higher $\lambda$ )

5 to 3

3 to 2

4 to 2 (lower $\lambda$ )

Note: Take into account that longer wavelength involves lower energy ( $E=\frac{hc}{\lambda}$ ).

3 0

3 years ago