First we need to find the limiting reactant. This is the reactant that is theoretically consumed completely first. Theoretical yield is how much product is produced given the limiting reactant is consumed and the percent yield is the fraction of the actual products produced and the theoretical products produced.
The limiting reactant is <span>Pb(<span>C2</span><span>H3</span><span>O2</span><span>)2.
The theoretical yield is </span></span>1.2373008g and you should get the Percent yield which is Actual yield/theoretical yield. <span />
Answer:
idk but i tryed
Explanation:
The simplest way to use the periodic table to identify an element is by looking for the element’s name or elemental symbol. The periodic table can be used to identify an element by looking for the element’s atomic number. The atomic number of an element is the number of protons found within the atoms of that element.
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
A cubic centimetre (or cubic centimeter in US English) (SI unit symbol: cm3; non-SI abbreviations: cc and ccm) is a commonly used unit of volume that extends the derived SI-unit cubic metre, and corresponds to the volume of a cube that measures 1 cm × 1 cm × 1 cm.
