<span>ideal gas law: PV = nRT so .....</span><span> V = PV/(RT) </span>
<span>
Initial number of moles of Cl, n = 0.943*5.11/(0.08206 × 286) mol = 0.2053 moles.
</span><span>
We know the molar mass of K (potassium) = 39.0 g/mol </span>
<span>sooo....
The Initial number of moles of K = 29.0 g/(39.0 g/mol) = 0.7436 moles</span>
<span>Find the balanced equation for the reaction : </span><span>2K + Cl2 → 2KCl </span>
<span>Mole ratio of K:Cl = 2:1 </span>
<span>So after the reaction, the amount of K needed = (0.2053 mol) × 2 = 0.4106 mol which is less than 0.7436 mol </span>
<span>
This means that K is in excess but Cl completely reacts. </span>
<span> So we know the mole ratio is Cl:KCl = 1 : 2
</span>
<span>Number of moles of Cl (completely) reacted = 0.2053 mol which means the n</span><span>umber of moles of KCl formed = (0.2053 mol) × 2 = 0.4106 mol </span>
<span>Molar mass of KCl = (39.0 + 35.5) g/mol = 74.5 g/mol </span>
<span>Mass of KCl formed = 0.4106 mol * 74.5 g/mol = 30.6 g</span>
Answer:
b. oxygen side being slightly negative and the hydrogen side being slightly positive.
Explanation:
The water molecule is a polar molecule, that is to say that its distribution of electronic density is different throughout the molecule.
In this way, in the water molecule there is a negative partial charge towards the oxygen atom and a positive partial charge towards the hydrogen atom.
This polar characteristic of the water molecule allows ions and other molecules to exhibit water solubility and is widely used in chemical reactions.
2.75 x 10^24
Hope this helped :)
Answer:
24.525 g of sulfuric acid.
Explanation:
Hello,
Normality (units of eq/L) is defined as:
Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:
Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:
Best regards.
D. Matter and energy are the same.
