Explanation:
Molar mass
The mass present in one mole of a specific species .
The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .
(a) S₈
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
Molar mass of S₈ = 8 * 32 g/mol. = 256 g/mol.
(b) C₂H₁₂
Molar mass of of the atoms are -
Hydrogen , H = 1 g/mol
Carbon , C = 12 g/mol
Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol
(c) Sc₂(SO₄)₃
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
oxygen , O = 16 g/mol.
scandium , Sc = 45 g/mol.
Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol
(d) CH₃COCH₃ (acetone)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol
(e) C₆H₁₂O₆ (glucose)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.
D.
This is self-regulation because when the population of the insects becomes too large, it regulates itself and starts to decrease due to a shortage of resources.
This is an incomplete question, here is a complete question.
Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.
CaCO₃, Ksp = 8.7 × 10⁻⁹
Answer : The solubility of CaCO₃ is, 
Explanation :
As we know that CaCO₃ dissociates to give
ion and
ion.
The solubility equilibrium reaction will be:

The expression for solubility constant for this reaction will be,
![K_{sp}=[Ca^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
Let solubility of CaCO₃ be, 's'




Therefore, the solubility of CaCO₃ is, 
Answer:
<h3>A-5N B-6N C-7N D-8N</h3>
Explanation:
i hope it helps ;)
Answer:
The question has some details missing. here are the details ; Given the following ;
1. 43.2 g of tablet with 20 cm3 of space
2. 5 cm3 of tablets weighs 10.8 g
3. 5 g of balsa wood with density 0.16 g/cm3
4. 150 g of iron. With density 79g/cm 3
5. 32 cm3 sample of gold with density 19.3 g/cm3
6. 18 ml of cooking oil with density 0.92 g/ml
Explanation:
<u>Appropriate for calculating mass</u>
32 cm3 sample of gold with density 19.3 g/cm3
18 ml of cooking oil with density 0.92 g/ml
<u>Appropriate for calculating volume</u>
5 g of balsa wood with density 0.16 g/cm3
150 g of iron. With density 79g/cm 3
<u>Appropriate for calculating density</u>
43.2 g of tablet with 20 cm3 of space
5 cm3 of tablets weighs 10.8 g