Answer:
The correct answer is C. element
Explanation:
The sample cannot be an element because an element - or <em>elemental substance</em> - cannot be decomposed into simpler substances. Thus, it cannot be composed by differents types of atoms. For example, an element is carbon (C).
As the sample contains <u>three types of atoms</u>, it can be a compound, a molecule or a mixture, because they can be composed by different types of atoms - of different chemical elements. For example, the sample could contain the element carbon (C) combined with other elements, for example oxygen (O) or hydrogen (H), amoing others.
CaCO3(s) ⟶ CaO(s)+CO2(s)
<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>
<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3,
therefore 0.0131 moles CO2 should also be formed.
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 </span>
Therefore:<span>
<span>% Yield: 0.53/.576 x100= 92 percent yield</span></span>
Answer:
19.3 L
Explanation:
V= n × 22.4
where V is volume and n is moles
First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01
n= m/ MM
n= 38/ 44.01
n= 0.86344012724
V= 0.86344012724 × 22.4
V= 19.3410588502 L
V= 19.3 L
Answer: this question is 3 days ago? Omg