Molar mass:
KCl = 74.55 g/mol
KClO3 = 122. 55 g/mol
<span>Calculation of the mass of KClO3 :</span>
<span>2 KClO3 = 2 KCl + 3 O2</span>
2* 122.55 g KClO3 ------------------ 2 * 74.55 g KCl
mass KClO3 ?? --------------------- 25.6 g KCl
mass KClO3 = 25.6 * 2 * 122.55 / 2 * 74.55
mass KClO3 = 6274.56 / 149.1
mass = 42.082 g of KClO3
Therefore:
1 mole KClO3 ---------------------- 122.55 g
?? moles KClO3 ------------------- 42.082 g
moles KClO3 = 42.082 * 1 / 122.55
moles KClO3 = 42.082 / 122.55
=> 0.343 moles of KClO3
Answer C
hope this helps!
Answer:
The atomic mass of carbon (C) is 12.0107 amu, so if you want to calculate the total mass in each molecule, you just need to multiply the number of carbon atoms in the substance by 12.017. In (a) there is one atom of C, (b) have also one atom of C, (c) have 12 atoms of C, and (d) have five atoms of C. Thus, the total mass (amu) of carbon is:
(a) 12.017 amu
(b) 12.017 amu
(c) 144.204 amu
(d) 60.085 amu
Alkali metals, alkali earth metals, transition metals, noble gases, halogens, and chalcogens
P₂O₅ is an empirical formula
<h3>Further explanation</h3>
Given
compounds
Required
an empirical formula
Solution
The empirical formula is the smallest comparison of the atoms forming the compound.
Molecular formulas are formulas that show the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
1. P₂O₅ : the ratio of compounds cannot be further reduced to whole numbers⇒empirical formula
2. P₄O₆ : molecular formula
(P₂O₃)₂=P₄O₆
P₂O₃ : empirical formula
3.C₂H₄ : molecular formula
(CH₂)₂=C₂H₄
CH₂ : empirical formula
4. C₃H₆ : molecular formula
(CH₂)₃=C₃H₆
CH₂ : empirical formula
Answer:
The answer is
<h2>
![2.143 \times {10}^{18} \: \: atoms](https://tex.z-dn.net/?f=%202.143%20%5Ctimes%20%20%7B10%7D%5E%7B18%7D%20%20%5C%3A%20%20%5C%3A%20atoms)
</h2>
Explanation:
To find the number of atoms given the number of moles we use the formula
N = n × L
where
N is the number of entities
n is the number of moles
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question
![n = 3.559 \times {10}^{ - 6} \: mol](https://tex.z-dn.net/?f=n%20%3D%203.559%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%206%7D%20%20%5C%3A%20mol)
Substitute the values into the above formula and solve
That's
<h3>
![N = 3.559 \times {10}^{ - 6} \times 6.02 \times {10}^{23} \\ = 2.1425 \times {10}^{18}](https://tex.z-dn.net/?f=N%20%20%3D%203.559%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%206%7D%20%20%5Ctimes%206.02%20%5Ctimes%20%20%7B10%7D%5E%7B23%7D%20%20%5C%5C%20%20%3D%202.1425%20%5Ctimes%20%20%7B10%7D%5E%7B18%7D%20)
</h3>
We have the final answer as
<h3>
![N = 2.143 \times {10}^{18} \: \: atoms](https://tex.z-dn.net/?f=N%20%20%3D%202.143%20%5Ctimes%20%20%7B10%7D%5E%7B18%7D%20%20%5C%3A%20%20%5C%3A%20atoms)
</h3>
Hope this helps you