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Bess [88]
4 years ago
9

Why is sodium more reactive than lithium​

Chemistry
2 answers:
creativ13 [48]4 years ago
7 0

Answer:  Atomic radius of sodium is greater than lithim,which allows sodium to loose electron easily than lithium. electropositivity and mettalic character increases as we move along the group or top to bottom . this is the reason why sodium is more reactive than lithium .

Explanation:

Elis [28]4 years ago
4 0

Answer:

Sodium is more reactive than lithium because as we move down a group it is easy to lose electrons as the number of shells increases and nuclear charge decreases. As valence electrons take part in chemical reaction and its easier to lose electrons as we move down a group chemical reactivity increases. Therefore, sodium is more reactive than lithium.

Explanation:

I gotcha I searched it up

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What is this. please respond asap. 2 and 3
kolbaska11 [484]

A solid stays the same shape. The atoms a close together and don't move that much. A liquid can form to any shape, the atoms bounce around a little bit, but not much. A gas has no shape, the atoms are few and far between.

A.) Liquid

B.) Solid

C.) Gas

5 0
3 years ago
3. Which two atoms could potentially form alloys?
Gennadij [26K]

Answer:

Tin and Copper

Explanation:

The two atoms that could potentially form alloys are the tin and copper atoms.

Alloys are mixtures formed between two or more metals.

The goal of forming alloys is to take properties of one metal and add to that of another metal so as to enhance both materials.

Alloys are only formed between metals.

The only metal pair given from the choices is that of tin and copper.

Therefore, the two atoms that could potentially form alloys is tin and copper.

8 0
3 years ago
MARKING BRAINLIEST :)
Reil [10]

Answer:

a

Explanation:

3 0
3 years ago
A 64.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 32.0 ml of koh at 25 ∘c.
aleksley [76]
Hello!

The reaction between HBr and KOH is the following:

HBr+KOH→H₂O + KBr

To calculate the amount of HBr left after addition of KOH, you'll use the following equations:

HBr_f=HBr_i-KOH=([HBr]*vHBr)-([KOH]*vKOH) \\  \\ HBr_f=(0,25M*0,64L)-(0,5M*0,32L)=0 mol HBr

That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be neutral, close to 7. 

Have a nice day!
3 0
3 years ago
Help this is due soon!!!
denis23 [38]

Answer:

not 100% sure but I think it's unbalanced chemical

6 0
4 years ago
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