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kap26 [50]
2 years ago
9

How many electrons will metals generally have in their outer shell

Chemistry
1 answer:
Goshia [24]2 years ago
3 0

Answer:

1 - 3

Explanation:

- Look to see where metals are on the periodic table then look at what group they are in. The group tells you the number of valence electrons. Ex. 1A has one valence electron.

- Hope this helped! If you need a further explanation please let me know.

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All organic molecules have what element in common
joja [24]

Every organic molecules/compound contains carbon (c).

Some other very abundant are hydrogen, nitrogen, oxygen, phosphorus, and sulfur.

I learned this with the acronym CHNOPS.

C - Carbon

H - Hydrogen

N - Nitrogen

O - Oxygen

P - Phosphorus

S - Sulfur

Hope this helps!


6 0
3 years ago
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Examples of a pure substances
DiKsa [7]

Answer:

Explanation:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.

3 0
2 years ago
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Please help ASAP I’ll make you u as brainlister please 100 points pelaseee
inessss [21]

Answer:

(C) im pretty sure is the answer

Explanation:

6 0
3 years ago
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How many atoms or moles of each element are on either side of the yield arrow? Is this equation balanced? 2Na+H2O---> NaOH+2H
Anton [14]
4 NA+4H2O —> 4NaOH+2H2

I hope that helped
4 0
3 years ago
A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First
bekas [8.4K]

Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0
3 years ago
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