Answer:
116.88g of table salt (NaCl) contains two formula units
Explanation:
Now,
We know that 1 formula unit of sodium chloride has a molar mass of 58.44g/mol
Hence;
Mass of 1 formula unit = 58.44g
Mass of x formula units = 116.88g
x = 116.88g * 1 formula unit/58.44g
x = 2 formula units
Therefore;
116.88g of table salt (NaCl) contains two formula units
Answer:
Explanation:
<em>Ferrous Sulphate</em>
<em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em>
<em>, Sulphur Dioxide </em>
<em> as well as Ferric Oxide </em>
<em>.</em>
<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>
<em>Hence,</em>
<em>Now,</em>
<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>
<em>1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em>
.
<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>
<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>
<em> </em>
<em>b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds </em><em>decompose </em><em>into smaller elements and compounds. Here, as Ferrous Sulphate </em><em>decomposes </em><em>into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is </em><em>Decomposition Reaction.</em>
The answer should be D all of the above
Answer:
The elements in__Group_ 0 of the Periodic Table are called the_noble__gases. They are generally __unreactive_. because they have a__full_outer shell of electrons. So they do not need to gain__lose_or share _electrons_ with other atoms.
Now ,
C + O2 → CO2
According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.
No of moles = mass of the substance/molecular mass of the substance.
In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.
No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.
No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles
Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.
Also number of moles of O2 = mass of O2÷ molar mass of O2
Substituting number of moles of O2 as 0.1 we get
mass of O2(x) = Number of moles of O2 × Molar mass of O2
Mass of O2 (x) = 0.1 × 32= 3.2 g
Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.
