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snow_tiger [21]
3 years ago
7

2. A/an _______ bond allows metals to conduct electricity. A. ionic B. metallic C. covalent D. atomic

Chemistry
2 answers:
lisov135 [29]3 years ago
7 0
The appropriate answer is b. A metallic bond allows metals to conduct electricity. Metallic bonds are formed by atoms of metals in which the outer electrons of the atoms from a common electron cloud. In metallic bonding the atoms are not used up in the actual bond but are shared so they are free to move about to conduct electricity. Electrons in the other bond types are held in place in the molecule an are not free to move about so they cannot conduct electricity.
stellarik [79]3 years ago
3 0
A metallic bond allows metals to conduct electricity. 
You might be interested in
If i have 340mL of a 1.5 M NaBr solution, What will the concentration be for 1000mL?
Gekata [30.6K]

Answer:

0.51M

Explanation:

Given parameters:

Initial volume of NaBr = 340mL

Initial molarity  = 1.5M

Final volume  = 1000mL

Unknown:

Final molarity = ?

Solution;

This is a dilution problem whereas the concentration of a compound changes from one to another.

In this kind of problem, we must establish that the number of moles still remains the same.

    number of moles initially before diluting = number of moles after dilution

Number of moles  = Molarity x volume

Let us find the number of moles;

          Number of moles  = initial volume x initial molarity

Convert mL to dm³;

                  1000mL  = 1dm³

                     340mL gives \frac{340}{1000}   = 0.34dm³

Number of moles  = initial volume x initial molarity  = 0.34 x 1.5 = 0.51moles

Now to find the new molarity/concentration;

               Final molarity  = \frac{number of moles}{Volume}    = \frac{0.51}{1}    = 0.51M

We can see a massive drop in molarity this is due to dilution of the initial concentration.

6 0
3 years ago
1.
german

Order of metals from least reactive to most reactive: B <C <A <D

<h3>Further explanation</h3>

Reducing agents are substances that experience oxidation  

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series  

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

Let's analyze the statement in the problem

I. Only A, C and D react with 1 mol/L HCl to give H₂(e)

M + HCl ⇒ MCl + H₂(MCl : alkali, MCl₂ : alkaline earth)

A, C and D can react with 1 mol / L HCl, meaning metals A, C and D are located to the left of element H (more reactive), and B in the right of element H

II. When A is added to solutions of the other metal ions, metallic B and C are  formed but not D.

This means that metal A is more reactive than metals B and C, while D is more reactive than A, so metal D is the most reactive

3 0
2 years ago
help please, How much water would you need to add to the 10 grams of sodium hydroxide to make 100 grams of the 10% solution?
jonny [76]

Answer:

90

Explanation:

gshhhshussussussjsuushhshjssuss

5 0
3 years ago
PLS ANSWER QUICKLY
vichka [17]
1. D
2. DR
3. SR
4. DR
5. C
6. SR
7. S
8. D
10. C
3 0
2 years ago
Would the following configuration represent
makvit [3.9K]

Answer:

Excited state

Explanation:

4 0
3 years ago
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