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-BARSIC- [3]
2 years ago
9

What is held together by a metallic bond?

Chemistry
1 answer:
Wewaii [24]2 years ago
5 0

Answer:

metallic bond, force that holds atoms together in a metallic substance. The atoms that the electrons leave behind become positive ions, and the interaction between such ions and valence electrons gives rise to the cohesive or binding force that holds the metallic crystal together.

Explanation:

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The rate of chemical reactions generally ____ as temperature is raised
Zepler [3.9K]

The rate of chemical reactions generally happen <em>faster</em> when the temperature is raised.

This happens because the reactant's molecules move faster when the temperature is raised. The molecules start to bounce around more, increasing the chance for the reaction to happen, or to increase the speed at which the reaction occurs. Hope this helped.

6 0
4 years ago
Read 2 more answers
a teacher uses a bow and arrow to demonstrate accuracy and precision. she shoots several arrows, aiming at the exact center of t
Vera_Pavlovna [14]
A teacher uses a bow and arrow to demonstrate accuracy and precision. she shoots several arrows, aiming at the exact center of the target each time. thedrawing below shows where her arrows hit the target.
<span> The statement  that best describes her shots is
</span><span>Her shots were neither accurate nor precise
</span>hope it helps
7 0
3 years ago
The pressure on 2.50 L of gas is changed from 4.5 atm to 2.2 atm. What is the new volume of gas?
swat32

Answer:

<h2>5.11 L</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{2.5 \times 4.5}{2.2}  =  \frac{11.25}{2.2}  \\  = 5.1136666...

We have the final answer as

<h3>5.11 L</h3>

Hope this helps you

8 0
3 years ago
How many moles of Al2O3 can be formed from 10.0 g of Al?
Nutka1998 [239]

Answer:

n Al=  10/27( mol)- >n Al2O 3 =5/27(mol)

Explanation:

3 0
3 years ago
Nuclear decay<br>27Al + He 》 39P<br>​
dybincka [34]

Answer:

_{13}^{27}\text{Al} + \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}

Explanation:

The unbalanced nuclear equation is

^{27}\text{Al} + \rm \text{He} \longrightarrow \, ^{31}\text{P}

We can insert the subscripts, because these are the atomic numbers of the elements

_{13}^{27}\text{Al} + \, \rm _{2}\text{He} \longrightarrow \, _{15}^{31}\text{P}

That leaves only the superscript of He to be determined,

The main point to remember in balancing nuclear equations is that the sums of the superscripts must be the same on each side of the equation.  

Then

27 + x = 31, so x = 31 - 27 = 4

Then, your nuclear equation becomes

_{13}^{27}\text{Al} + \, \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}

6 0
3 years ago
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