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3 years ago

Of the elements fe li te u and he which are considered group b elements

Chemistry

2 answers:

kramer3 years ago

7 0

Only fe (iron) is in the group b of elements

givi [52]3 years ago

5 0

<u>Answer:</u> Iron is considered as a Group B element.

<u>Explanation:</u>

Mendeleev divided the periodic table into 8 groups, each divided into two sub-groups named as A and B. He arranged the elements according to their atomic masses.

Group A elements are considered as the elements which are present in s-block and p-block.

Group B elements are the transition elements. Transition elements are the elements that belong to d-block.

From the given elements:

Lithium (Group 1-A) and helium (Group VIII-A) are s-block elements.

Tellurium is a p-block (Group VI-A)element.

Uranium is a f-block element.

Iron is the d-block (Group VIII-B)element (transition element)

Hence, iron is considered as a Group B element.

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Distinguish between atom molecules, and ions

NNADVOKAT [17]

Explanation:

Atoms are the smallest unit of matter that can't be broken down chemically. Molecules are groups of two or more atoms that are chemically bonded. Ions are atoms or molecules that have gained or lost one or more of their valence electrons and therefore have a net positive or negative charge.

7 0

3 years ago

How could you engineer a device that produces beneficial friction and heat?

nlexa [21]

Answer:

Explanation:

That's why rubbing your hands together makes them warmer. ... Friction causes the molecules on rubbing surfaces to move faster, so they have more energy. This gives them a higher temperature, and they feel warmer. Heat from friction can be useful.

3 0

3 years ago

A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o

yaroslaw [1]

<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

Mass of the unknown metal sample as 58.932 g
Initial temperature of the metal sample as 101°C
Final temperature of metal is 23.68 °C
Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

Therefore; mass of pure water is 45.2 g
Initial temperature of water = 21°C
Final temperature of water is 23.68 °C
Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

= 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

= 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q = m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

= 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

= 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>

We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.

Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

= 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0

3 years ago

two students are working together to build a birdhouse. Student 1 applies a force of 10 N to a wooden board in order to slide it

Ganezh [65]

Answer:

Explanation:

if there is 10n and the friction is 4n on the table then it is 6n.

i hope this helps you! :)

5 0

3 years ago

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical

Katarina [22]

Answer:

empirical formula: $H_2SO_4$

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02 : 1.02/1.02 : 4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: $H_2SO_4$ .

7 0

3 years ago