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BartSMP [9]
3 years ago
11

A ___________________ is a combination of two or more atoms that are held together by covalent bonds.

Chemistry
2 answers:
Andre45 [30]3 years ago
3 0
The answer is Molecule
frosja888 [35]3 years ago
3 0
The answer is molecules.
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What volume in liters of chlorine gas would be fully required to react to 125 g of Nal
Ivahew [28]

Volume of Cl₂ = 9.34 L

<h3>Further explanation</h3>

Reaction

2NaI + Cl₂ ⇒ 2NaCl + I₂

mol NaI (MW=149,89 g/mol) :

\tt \dfrac{125}{149,89}=0.834

mol Cl₂ : mol NaI = 1 : 2

mol Cl₂ :

\tt \dfrac{1}{2}\times 0.834=0.417

Assumptions at STP(1 mol=22.4 L) :

\tt 0.417\times 22.4~L=9.34~L

4 0
3 years ago
How many nonbonding electron pairs are there in the lewis structure of the peroxide ion
DENIUS [597]

Answer:

There are 6 non bonding pair of electrons present in Lewis structure of peroxide ion.

Explanation:

Ionic formula of peroxide ion is O_{2}^{2-}

Total number of valence electron present in peroxide ion is 14 (12 electrons come from two oxygen atoms and another 2 electrons come from negative charges).

After fulfilling octet rule, Lewis structure of peroxide shows that each oxygen atom contains three non bonding electron pairs and a covalent bond.

Hence there are 6 non bonding pair of electrons present in Lewis structure of peroxide ion.

Lewis structure has been given below.

8 0
3 years ago
What temperature should it be to do a baking soda and vinegar mixture​
Leokris [45]

Answer:

16 degrees Celsius to 12 degrees to Celsius in 10 seconds

Explanation:

4 0
3 years ago
How are the elements arranged in the modern periodic table g?
Yuri [45]
Rows and numbers and periods 
3 0
4 years ago
Uranium has three common isotopes. If the abundance of 234U is 0.0054%, the abundance of 235U is 0.7204% and the abundance of 23
OlgaM077 [116]

Answer:

237.9 amu

Explanation:

Given data:

Abundance of U-234 = 0.0054 %

Abundance of  U-235 = 0.7204%

Abundance of  U-238 = 99.2742%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass  / 100

Average atomic mass = (0.0054×234)+(0.7204×235) +(99.2742×238)  /100

Average atomic mass =  1.2636+ 169.294+ 23627.2596 / 100

Average atomic mass = 23797.8172/ 100

Average atomic mass = 237.9 amu.

7 0
3 years ago
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