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denpristay [2]
2 years ago
10

How would you draw the ionic compound Beryllium Arsenide lewis structure?

Chemistry
1 answer:
astra-53 [7]2 years ago
8 0

Answer:

[Na]^+ [Cl]^-

Explanation:

Lets say its sodium, its number of electrons is 11, but when its stable (an ion), it is 10. and chloride, number of electrons is 17, but when its stable (an ion) it is 18. So the lewis structure for that is, remember with the straight brackets (not sure what it's called, but you know what I mean I guess) its this one: [ ]

Sodium will be + because it has more protons (11-10 = +1), and chloride will be - because it gained an electron, so has more electrons than protons (17-18 = -1)

So the lewis structure would be:

[Na]^+  for sodium

and

[Cl]^- for chlorine

Sodium chloride:

[Na]^+ [Cl]^-

Also just to add, only 1 of each atom (Na and Cl) was needed for the bonding, but if let's say example; 2 Cl was needed to bond with sodium, there would be 2 Cl (same) and 1 Na.

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1. Describe how SHAPE can change during a collision. Be specific with your evidence and add detail to your answer.
a_sh-v [17]

Answer:

Hey

of course, the damage of a collision depends upon how fast to objects are moving at each other and how strong they are. If you have two tanks moving at each other 2 miles per hour it will be very little damage and the ->shape<- will not change much, maybe a dint or two. But if two balloons filled with water are moving at each other 5000 mph they will completely evoporate in a burst of light, and their ->shape<- will change very much. This is how shape and motion are related.

Hope it helped

spiky bob your answerer

4 0
3 years ago
A 58.5 g sample of glass is put into a calorimeter (see sketch at right) that contains 250.0 g of water. The glass sample starts
Ket [755]

Answer:

The specific heat capacity of glass is 0.70J/g°C

Explanation:

Heat lost by glass = heat gained by water

Heat lost by glass = mass × specific heat capacity (c) × (final temperature - initial temperature) = 58.5×c×(91.2 - 21.7) = 4065.75c

Heat gained by water = mass × specific heat capacity × (final temperature - initial temperature) = 250×4.2×(21.7 - 19) = 2835

4065.75c = 2835

c = 2835/4065.75 = 0.70J/g°C

5 0
3 years ago
Read 2 more answers
What's it called when an electron has absorbed alot of energy?​
emmainna [20.7K]

Answer:

Excited state

Explanation:

Excited state is the name of the state when electrons absorb energy and move to a higher energy level.

Hope this helps!

8 0
3 years ago
1. Compare masses: a) 0,4mol CO₂ and 0,6mol H₂O ; b) 0,135mol H₂SO₄ and 0,5mol HCI.
bixtya [17]

Answer:

you can now deduct which one is greater or smaller and by how much.

Explanation:

no of moles= mass/molar mass

1ai) 0.4 = m/ ( 12 + (16*2)

m= 0.4* 44

m= 17.6g

ii) 0.6= m/( 2*1 + 16)

m= 0.6 *18

m= 10.8g

b) 0.135 = m/ ( 2*1 +32 + (16*4)

m= 0.135* 98

m= 13.23g

ii) 0.5= m/ (1+35.5)

m= 0.5*36.5

m= 18.25g

2. Avogadro's Number = 6.02×10²³

1 mol of any element= 6.02×10²³ particles

a) 0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules

ii) 0.3 mol of CO2= (6.02×10²³) * 0.3

= 1.806 × 10²³ molecules

Ans: 0.3 mol of CO2

bi) 0.25 mol of HCl= (6.02×10²³) * 0.25

= 1.505 × 10²³ molecules

bii) - find the no of moles first:

no of moles= mass/molar mass

n= 3.4g/ 34g →mr of H2S in g=2+32= 34g

n= 0.l mol

- use the Avogadro Number.

0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules.

biii) here you're given the density, use it to find the mass of acetic acid.

ρ = 1049 g/ml

ρ = m/v, where v=5 ml

1049 = m/ 5

m= ρ*v

m= 1049*5

m= 5245 g

• convert this into moles.

mr of CH3COOH= 12 + 3+ 12+ 16+ 16+ 1

= 60

mr in g = 60g

n= m/mr

n= 5245/ 60

n= 87. 41666...

n= 87.4 moles

•using Avogadro's Number:

87.4 moles of acetic acid=(6.02×10²³)*87.4

= 2.25148* 10²⁵

= 2.25 * 10²⁵ molecules

thus, the ans for this is 5 ml of acetic acid.

7 0
2 years ago
Let's say you have a mixture for which you want to check the composition. Unfortunately, after quite a while, and several TLC an
KIM [24]

Answer:

mmmmmmmmmmmmmm

Explanation:

7 0
3 years ago
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