Answer:
The concentration in mol/L = 4.342 mol/L
Explanation:
Given that :
mass of sodium chloride = 25.4 grams
Volume of the volumetric flask = 100 mL
We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol
and number of moles = mass/molar mass
The number of moles of sodium chloride = 25.4 g/58.5 g/mol
The number of moles of sodium chloride = 0.434188 mol
The concentration in mol/L = number of mol/ volume of the solution
The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L
The concentration in mol/L = 4.34188 mol/L
The concentration in mol/L = 4.342 mol/L
Answer:
C. It is reactive because it must gain two electrons to have a full outermost energy level.
Step-by-step explanation:
The electron configuration of this element ends in 2s²2p⁴.
A filled energy level would end in 2s²2p⁶.
The element will be reactive, because it must gain electrons to have a full energy level, and it needs two more electrons to do this.
<span>In chemistry, a catalyst can speed up the reaction (or make it initiate easier) by altering the activation energy, lowering it enough to allow the reactants to react more easily. Some negative catalysts or inhibitors can do the same by increasing the activation energy.
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Answer:
32.7 g of Zn
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Zn + 2HCl —> ZnCl₂ + H₂
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂
Next, we shall determine the number of mole of Zn required to produce 0.5 mole of H₂. This can be obtained as follow:
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂.
Therefore, 0.5 mole of Zn will also react to produce to 0.5 mole of H₂.
Thus, 0.5 mole of Zn is required.
Finally, we shall determine the mass of 0.5 mole of Zn. This can be obtained as follow:
Mole of Zn = 0.5 mole
Molar mass of Zn = 65.4 g/mol
Mass of Zn =?
Mass = mole × molar mass
Mass of Zn = 0.5 × 65.4
Mass of Zn = 32.7 g
Thus, 32.7 g of Zn is required to produce 0.5 mole of H₂.
CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).
