Which of the following is a solution of the equation 16a = 4?
Which of the following is a solution of the equation 1 6 a = 4? => 4 or - 4 or 1 4 or - 1 4..
Solution
The fuel efficiency can be predicted from the amount of heat evolved for every gram of fuel consumed. (i) The combustion of methane is as follows: D H c = 1559.7kJmol - 1 Molar mass of C 2 H 6 = 30..
The fuel efficiency can be predicted from the amount of heat evolved for every gram of fuel consumed. (i) The combustion of methane is as follows: D H c = 1559.7kJmol - 1 Molar mass of C 2 H 6 = 30..Purification of Colloidal Sols
The colloidal solutions prepared by various methods usually contain electrolytes and other soluble substances as impurities. These impurities if not removed can destabilize the sols. Impurities are removed by the following methods: 1) Dialysis 2) Ultra-filtra..
Solution
The fifth period begins with the filling of '5s' orbitals and continues till the filling of sixth energy level (6s) starts. The subshells which follow '5s' are, 4d, 5p, 6s, Thus, the elements which involve filling of 5s, 4d, 5p, sub shells are acc..
Solution
= 0.05608 = 0.0561 The number of significant figures in the term 3.24 is 3, therefore the result should have 3 significant figures. Therefore, the correct answer is 0.0561. The number after zero is 8 and therefore it is rounded off to one. = 0.2615 x 10 - 4 = 0.3 x 10 - ..
= 0.05608 = 0.0561 The number of significant figures in the term 3.24 is 3, therefore the result should have 3 significant figures. Therefore, the correct answer is 0.0561. The number after zero is 8 and therefore it is rounded off to one. = 0.2615 x 10 - 4 = 0.3 x 10 - ..Solution
The degree of hardness of water is defined as the number of parts by mass of calcium carbonate, equivalent to various calcium and magnesium present in one million parts of mass of water. It is expressed in ppm. Now, 10 6 g of water contains CaCO 3 = 44 mg By definition, ..
The degree of hardness of water is defined as the number of parts by mass of calcium carbonate, equivalent to various calcium and magnesium present in one million parts of mass of water. It is expressed in ppm. Now, 10 6 g of water contains CaCO 3 = 44 mg By definition, ..Solution
The above results are written as follows: So, The empirical formulae of the compound = C 1 H 4 or CH 4 The empirical formula mass = (1 x 12) + (4 x 1) = 12 + 4 = 16 amu Molecular mass (given) = 16 amu Therefore, ..
The above results are written as follows: So, The empirical formulae of the compound = C 1 H 4 or CH 4 The empirical formula mass = (1 x 12) + (4 x 1) = 12 + 4 = 16 amu Molecular mass (given) = 16 amu Therefore, ..Solution
Data given: 92.20% of silicon of mass 27.98 amu 4.7% of silicon of mass 28.98 amu 3.1 % of silicon of mass 29.97 amu Average atomic weight = 25.80 + 1.36 + 0.93 = 28.09 amu. Average atomic weight of silicon is 28.09 amu.11. One million silver atoms weigh h..
Data given: 92.20% of silicon of mass 27.98 amu 4.7% of silicon of mass 28.98 amu 3.1 % of silicon of mass 29.97 amu Average atomic weight = 25.80 + 1.36 + 0.93 = 28.09 amu. Average atomic weight of silicon is 28.09 amu.11. One million silver atoms weigh h..Solution
Mass number of 1 4 C is 14. So 14 g of 1 4 C contain 6.023 x 10 2 3 atoms of 1 4 C = 3.012 x 10 2 0 atoms Number of neutrons in 1 atom of 1 4 C = 14 - 6 = 8 Number of neutron..
Mass number of 1 4 C is 14. So 14 g of 1 4 C contain 6.023 x 10 2 3 atoms of 1 4 C = 3.012 x 10 2 0 atoms Number of neutrons in 1 atom of 1 4 C = 14 - 6 = 8 Number of neutron..Solution:
The given problem is solved in the following table. Empirical formula = C 1 H 2 O 1 = CH 2 O Empirical formula mass = (12 x 1) + (1 x 2) + (16 x 1) = 30 Molecular mass = 180 (given) Ratio of the molecular mass to empirical formula ma..
The given problem is solved in the following table. Empirical formula = C 1 H 2 O 1 = CH 2 O Empirical formula mass = (12 x 1) + (1 x 2) + (16 x 1) = 30 Molecular mass = 180 (given) Ratio of the molecular mass to empirical formula ma..See what our Users say :
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