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sasho [114]
3 years ago
9

Which model shows 6 electrons in the outer shell of the atom?

Chemistry
2 answers:
AVprozaik [17]3 years ago
7 0
D. Model number 2

Hope this helps!! (:
fgiga [73]3 years ago
5 0

Answer:

2 has 6 on the outer shell

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Please help!
ANEK [815]

Answer:

A- Physical, B- Chemical, C- chemical, D- Physical

Explanation:

A is physical because you can see it changing its form is changing.

B is Chemical because a new substance is formed creating the orange color of rust.

C is a Chemical reaction because it is being broken down so the banana itself is changing not just how we see it.

D is physical because we are just changing the shape/ size of the item, not anything to do with its substances.

8 0
2 years ago
Read 2 more answers
Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s)
klemol [59]

Answer:

a) 2.01 g

Explanation:

  • Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:

  • 0.0302 mol AgNO₃ * \frac{1molNa_2CO_3}{2molAgNO_3}  = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

  • 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:

  • 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).

8 0
3 years ago
Select the correct answer from each drop-down menu.
seraphim [82]

1. protons and neutrons

2. electrons

5 0
3 years ago
Read 2 more answers
A rock contains 0.623 mg of 206Pb for every 1.000 mg of 238U present. Assuming that no lead was originally present, that all the
maxonik [38]

Answer:

t = 3,496x10⁹ years

Explanation:

The decay of ²³⁸U is:

²³⁸U → ²⁰⁶Pb + 8He + 6e⁻

Moles of ²⁰⁶Pb presents in 0,623mg are:

0,623x10⁻³g×(1mol / 206g) = 3,02x10⁻⁶ moles of ²⁰⁶Pb.

These moles are equals to moles of ²³⁸U before decay, that means, 3,02x10⁻⁶ moles²³⁸U

In grams:

3,02x10⁻⁶ moles²³⁸U× (238g / 1mol) = 7,20x10⁻⁴ g ²³⁸U = 0,720 mg²³⁸U

That means initial ²³⁸U was 1,000mg + 0,720mg =<em> 1,720mg</em>

Applying the formula:

ln (N₀/N) t₁₂ = t ln2

Where N₀ is initial amount of uranium (1,720mg), N is concentration of uranium (1,000mg),  half-life time is a constant (t₁₂= 4,468x10⁹ years) and t is the time transcurred for the reaction. Replacing:

ln(1,720/1)*4,468x10⁹ years = t ln2

<em>t = 3,496x10⁹ years</em>

<em></em>

I hope it helps!

6 0
3 years ago
Use the name to write the formula for the following ionic compound: scandium (III) hydroxide
asambeis [7]

Answer:

.

Explanation:

.

6 0
2 years ago
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