The amount of magnesium nitrate are formed from the 5.5 moles of the nitric acid in grams are 407.8 g.
The reaction is as :
2HNO₃ + Mg(OH)₂ ----> Mg(NO₃)₂ + 2H₂O
The number of moles of nitric acid = 5.5 moles
2 moles of the nitric acid = 1 mole of the magnesium nitrate
The number of moles of nitric acid = 5.5 / 2 mol
The number of moles of nitric acid = 2.75 mol
The grams of the magnesium nitrate = moles × molar mass
The grams of the magnesium nitrate = 2.75 mol × 148.3 g/mol
The grams of the magnesium nitrate = 407.8 g
Therefore, the mass of the magnesium nitrate is 407.8 g.
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Answer: The two isotopes of boron differs in mass number rather than atomic number.
Explanation:
Isotopes are defined as the chemical species that have same atomic number but differ in their mass number.
Boron is the 5th element of the periodic table and have 2 naturally occurring isotopes. The isotopes are : B-10 and B-11
For isotope:
Percentage abundance of this isotope = 20 %
Atomic number = 5
Mass number = 10
For isotope:
Percentage abundance of this isotope = 80 %
Atomic number = 5
Mass number = 11
Hence, the two isotopes of boron differs in mass number rather than atomic number.
(2) 2 (4) 18
Answer: (3) 15
Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.
2. Calculate the moles of O in the sample.
Answer:
The answers are
C = 0.0895 moles
1. 0.02975 moles of N
2. 0.0299 moles O
Explanation:
Mass of sample = 2.18 g
Combustion yields 3.94 g of CO₂ and 1.89 g of H₂O
1.23 g of sample contains 0.235 g of N
3.94 g of CO₂ which contains one mole of C and two moles of O
Molar mass of CO₂ = 44 mass fraction of C = 12/44 = 0.273, Therefore mass of carbon in 2.18 g = 0.273×3.94 = 1.075 g
number of moles of C in sample = mass/(molar mass) = 1.075 g/(12.01 g/mol) = 0.0895 moles of C
(1) Mass of nitrogen, in 1.23 g of sample = 0.235
therefore mass in 2.18 g =0.235×2.18/1.23 = 0.4165 g
= 0.02975 moles of N
(2) To find the number of moles of oxygen in the sample we note that total mass of sample = 2.18 g containing only nitrogen, hydrogen, oxygen and carbon
Mass of hydrogen can be fround from 2/18×1.89 = 0.21 g
Mass of oxygen therefore = 2.18 -(0.21+0.4165+1.075) = 0.4785
number of moles = 0.475/16 = 0.0299 moles
b. secondary succession has soil
c. primary succession has soil
d. primary succession has water
When the substance was tested in the lab it was seen that it maintains a unique shape, which is the differentiating property of solids. Also, they are very good conductors of electricity. Solid contains electrons, protons and neutrons this is a common thing which is found in every state of matter. This all details confirms that it will be solid.