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Elis [28]
3 years ago
13

Another word for "basic" is?

Chemistry
2 answers:
irakobra [83]3 years ago
8 0
Simple...I looked this up on google..hope this helps
natita [175]3 years ago
5 0

Answer: Fundamental, Key, vital, crucial

Explanation:

You might be interested in
Why do minerals with metallic bonds conduct electricity so well?
Roman55 [17]

Answer and Explanation:

Because metallic bonding involves delocalized electrons. It is described as a "<em>sea of electrons</em>", because the electrons are not confined around the nucleus of metal atoms, but they are delocalized: thay can be located in one nucleus and then in another neighbor atom. Thus, the electrons have more freedom to move from one part of the metal to another and electricity is well conducted.

8 0
3 years ago
What is the percent composition by mass of oxygen in ca(no3)2 (gram-formula mass = 164 g/mol)?a.9.8%b.29%c.48%d.59%?
Mila [183]
Percent composition by mass of oxygen =
((16.0*6)/(40.1+2*(14.0+16.0*3)))*100%
= 58.5%
therefore, the answer is D
8 0
3 years ago
Read 2 more answers
The atomic symbol represents the _____ of the element.
Anarel [89]

Protons

Explanation: the atomic number is the same as the number of protons

4 0
3 years ago
I have no idea what to do please help me quickly!
Irina18 [472]

Answer:

The farther away the planet the slower the revolution around the earth. the closer the faster.

Explanation:

its like a tetherball pole when it wraps around it gets closer and spins faster and faster untill it stops. Brainliest?

6 0
3 years ago
Read 2 more answers
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
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