Answer:
[Na⁺] = 1.99 M
Explanation:
Na₂S is a ionic salt that can be dissociated.
Dissociation equation is:
Na₂S → 2Na⁺ + S⁻²
1 mol of sodium sulfide can give 2 moles of sodium cation.
We convert moles of salt: 78 g . 1mol / 78.06 g = 0.999 moles
As ratio is 1:2, after dissociation we have (0.999 . 2) = 1.998 moles of Na⁺
Molarity is a type of concentration.
It indicates moles of solute in 1 L of solution and in this case, we have 1 L as final voulme.
Moles of Na⁺ are 1.998 moles. Then molarity (mol/L) is:
M =1.99 mol/L
Answer:1. 5
Explanation: the element is boron and the atomic number is 5
An oxide of nitrogen contains 30.45 mass % N, if the molar mass is 90± 5 g/mol the molecular formula is N₂O₄.
<h3>What is molar mass?</h3>
The molar mass of a chemical compound is determined by dividing its mass by the quantity of that compound, expressed as the number of moles in the sample, measured in moles. A substance's molar mass is one of its properties. The compound's molar mass is an average over numerous samples, which frequently have different masses because of isotopes.
<h3>How to find the molecular formula?</h3>
The whole-number multiple is defined as follows.
Whole-number multiple = 
The empirical formula mass is shown below.
Mw of empirical formula = Mw of N+ 2 x (Mw of O)
= 14.01 g/mol + 2 x (16.00 g/mol)
= 46.01 g/mol
With the given molar mass or the molecular formula mass, we can get the whole-number multiple for the compound.
Whole-number multiple =
≈ 2
Multiplying the subscripts of NO2 by 2, the molecular formula is N(1x2)O(2x2)= N2O4.
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Answer:
7 M
Explanation:
The molarity is defined as "moles per liter." Here, it is ...
(14 moles)/(2 liters) = 7 moles/liter = 7 M
Answer:
A)Chlorine and Bromine:
They are both non metal hence they form a covalent bond due to covalent bonding.
B)Potassium and Helium:
Helium ion has a small cationic radius and distorted by the potassium ion due to polarization.
C)Sodium and Lithium:
Both are metals hence they form a metallic bond since they share electrons to the electron cloud.