The required formula of hydrate is MgSO₃.6H₂O.
<h3>How do we calculate the formula of hydrate?</h3>
The number of moles of water per mole of anhydrous solid (x) will be computed by dividing the number of moles of water by the number of moles of anhydrous solid (x) to find the hydrate's formula.
Moles will be calculated as:
n = W/M, where
- W = given mass
- M = molar mass
Moles of MgSO₃ = 0.737g / 104.3g/mol = 0.007mol
Moles of H₂O = 0.763g / 18g/mol = 0.04 mol
Number of H₂O molecule = 0.04/0.007 = 5.7 = 6
So formula of hydrate is MgSO₃.6H₂O.
Hence required formula of hydrate compound is MgSO₃.6H₂O.
To know more about hydrate compound, visit the below link:
brainly.com/question/22411417
#SPJ1
Answer:
4
Explanation:
Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.
The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
