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Ugo [173]
3 years ago
6

How would I complete and balance the equation Mg + Br2 >

Chemistry
1 answer:
Katyanochek1 [597]3 years ago
3 0
Hope this helps you.

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Please let me know if this is correct?
siniylev [52]

Answer: Yes! you're all good. Alkali metals in group 1 are the most metallic :)

6 0
3 years ago
15g baking soda is dissolved in 100 mL water. The solute of the
Mumz [18]

Answer:

Baking soda

Explanation:

5 0
2 years ago
Pls hlp me with these questions: brainliest, ratting, thx, etc.
daser333 [38]
1 elements
2 they have same number of valence electrons
3 period
4 0
2 years ago
Please help.
mojhsa [17]

Explanation: An element is represented in the form of _{A}^{Z}\textrm{X} where,

X = Symbol of the element

A = Atomic Mass of the element X

Z = Atomic Number of element X

Hence, For the element Justwondoricium,

Symbol = Jw

Atomic number = 120

Atomic Mass = 224

Now, Atomic number = Number of electrons = Number of Protons

Number of electrons = 120

Number of Protons = 120

and Atomic Mass = Number of neutrons + Number of protons

Number of neutrons can be calculated as we know atomic mass and number of neutrons, Putting the numbers we get

Number of neutrons = 224 - 120 = 104

The nearest noble gas to the element having atomic number 120 is Oganesson (Og), which has an atomic number of 118, so the next two electrons will be filled in the 8s orbital.

Electronic Configuration of Jw is [Og]8s^2

This electronic configuration lets us know about the location of the element in periodic table.

As the electron is entering the 8th shell, it belongs to the 8th period and as the last electron enters the s-orbital, it belongs to the S-block of the periodic table.

When the s-electrons are 1, it belongs to Group 1.

When the s-electrons are 2, it belongs to Group 2.

As this element has 2 s-electrons, it belongs to the Group 2.

8 0
3 years ago
If 11.0 g of ccl3f is enclosed in a 1.1 −l container, will any liquid be present?if so, what mass of liquid?
Anastasy [175]
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law 
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>

<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>

<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol.  </span>

<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>

<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
8 0
2 years ago
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