<span>write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
You are given everything needed to calculate
q=heat transfer=2.2*10^2,
H3PO4 moles= 1.5*10^-3,
NaOH moles=5.0*10^-3
equation is deltaHneutraliztion=q/Moles of limiting reagent
H3PO4 is limiting reagent because lowest moles, and is used up first
Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj,</span>
Chemical- it produces ammonia.
Answer:
Carbon - 14
Oxygen - 16
Nitrogen - 15
Sulphur - 16
Explanation:
The question above is related to the "Periodic Table of Elements" which shows the proper arrangement of elements in a table. Every element falls on a<em> group/family</em> within the table. Each group has its own number, and the table has a total of<u> 18 groups</u><em> (from left to right). </em>They are classified according to <em>similarities in their characteristics</em>. For example, group 1 is composed of <em>alkali metals</em> while group 2 is composed of<em> alkali earth metals</em>.
First we have to refer
to the reaction between the acid and the base: <span>
H2SO4 + 2 NaHCO3 ---> 2 H2O + 2 CO2 + Na2SO4
From this balanced equation we can see that for every 1 mol
of acid (H2SO4), we need 2 mol of base (NaHCO3) to neutralize it. Given 28 ml
of 5.8 M acid, we need to find out how many mols of acid that is:
<span>28mL * (1L/1000mL) * 5.8 mol/L = 0.1624 mol H2SO4</span></span>
<span>
Since we need 2 mol of base per mol of acid, we need:</span>
<span> 2*0.1624 mol = 0.3248 mol NaHCO3 </span><span>
MolarMass of NaHCO3 is 84.01 g/mol
<span>0.3248 mol*(84.01g/mol) = 27.29 g NaHCO3</span></span>
