Answer:
M = 0.050 M
Explanation:
WE have an acid base reaction between NaOH and HCl. The reaction is as follow:
HCl + NaOH -------> NaCl + H₂O
As we can see, we have a balanced equation already, and we can also see that the mole ratio between the HCl and NaOH is 1:1, therefore we can use the following expression:
M₁V₁ = M₂V₂ 1: acid; 2: base
This expression is used when the reactions has reached the equivalence point. As we want to know the concentration of the base, we just solve for M₂ above so:
M₂ = M₁V₁ / V₂
All we have to do now is replace the given values and solve for the concentration:
M₂ = 0.15 * 27 / 81
<h2>
M₂ = 0.050 M</h2><h2>
This is the concentration of the base</h2>
17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

NaClO₄ is a strong electrolyte that dissociates according to the following equation.
NaClO₄ ⇒ Na⁺ + ClO₄⁻
The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
You can learn more Avogadro's number here: brainly.com/question/13302703
1) H2O is able to dissolve both polar molecules and non polar ones
2) due to its extreme polarity it can even dissolve some I onic compounds
3 the h2o molecule itself is small in size
Answer:
When dry ice is exposed to warm air, a gas called carbon dioxide is produced through a process called sublimation
Explanation:
Answer : The average atomic mass of an element X is, 63.546 amu
Solution : Given,
Mass of isotope X-63 = 62.9296 amu
% abundance of isotope X-63 = 69.15% = 0.6915
Mass of isotope X-64 = 64.9278 amu
% abundance of isotope X-64 = 30.85% = 0.3085
Formula used for average atomic mass of an element X :
Therefore, the average atomic mass of an element X is, 63.546 amu