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nadya68 [22]
3 years ago
15

A student uses 5g of carbon to extract copper metal from its ore:

Chemistry
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

drt4gxhknwo

Explanation:

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Which of the following trends can be identified on the periodic table?
Jet001 [13]

Answer:

D) atomic radii increase from top to bottom of a group

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also  decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

Other options are incorrect because,

A) atomic radii increase from left to right across the period

Correct = atomic radii decreases from left to right across the period

B) ionization energy increases from top to bottom within a family

Correct =  ionization energy decreases from top to bottom within a family

C) electronegativity decreases from left to right across a period

Correct = electronegativity increases from left to right across a period

8 0
4 years ago
I really need help..please and thank you.
Elenna [48]

What is the question you need answered

8 0
3 years ago
A 47.1 g sample of a metal is heated to 99.0°C and then placed in a calorimeter containing 120.0 g of water (c = 4.18 J/g°C) at
wlad13 [49]

Answer : The metal used was iron (the specific heat capacity is 0.44J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

C_1 = specific heat of metal = ?

C_1 = specific heat of water = 4.18J/g^oC

m_1 = mass of metal = 47.1 g

m_2 = mass of water = 120 g

T_f = final temperature of water = 24.5^oC

T_1 = initial temperature of metal = 99^oC

T_2 = initial temperature of water = 21.4^oC

Now put all the given values in the above formula, we get

47.1g\times c_1\times (24.5-99)^oC=-120g\times 4.18J/g^oC\times (24.5-21.4)^oC

c_1=0.44J/g^oC

Form the value of specific heat of metal, we conclude that the metal used in this was iron.

Therefore, the metal used was iron (the specific heat capacity is 0.44J/g^oC).

3 0
3 years ago
Plsss help <br> will mark BRAINLIEST
satela [25.4K]

Answer:

c

Explanation: just a chemachal

5 0
3 years ago
Quick help me please
Nataly [62]
A or b I think it’s b
5 0
3 years ago
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