Answer:
a. Are miscible because each can hydrogen bond with the other.
Explanation:
Both ethanol and water are miscible. The reason why they can both mix freely is due to the hydrogen bonds that will form between their molecular structure.
Hydrogen bonds are special dipole-dipole attraction between polar molecules in which hydrogen atoms are directly joined to an electronegative atom.
Ethanol has an hydroxyl group which will bond to form an intermolecular bond with the oxygen and hydrogen on the water molecule. This attraction makes them miscible.
Answer:
D
Explanation:
As bioindicators are the organism that indicate or monitor the health of the environment
Answer:
Option B. is the right answer.
Explanation:
Before developing a new technology, an engineer has to evaluate the advantages and disadvantages of the chemical used in order to clean water bodies. If the chemical has more disadvantages as compared to advantages, so its usage will be avoided while if the chemical does not harm the marine organisms of the ocean so it can be used for the purpose of cleaning. So we can say that first the engineer has to study the chemicals.
From the reactions, 1.04 g of H2 and 7.995 g of aluminum phosphate is produced.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with the amount of substances that participates in a reaction.
For reaction 1;
Mg + 2HCl → MgCl₂ + H₂
Number of moles of Mg reacted = 12.5 g/24g/mol = 0.52 moles
If 1 mole of Mg produced 1 mole of H2
0.52 moles produces 0.52 moles of H2
Mass of H2 = 0.52 moles * 2 g/mol = 1.04 g
For reaction 2;
2Li3PO4 + Al2(SO4)3 → 3Li2SO4 + 2AIPO4
Number of moles of lithium phosphate = 7.5 g/116 g/mol = 0.065 moles
2 moles of Li3PO4 produced 2 moles of AIPO4
0.065 moles of Li3PO4 produced 0.065 moles of AIPO4
Mass of AIPO4 = 0.065 moles * 123 g/mol = 7.995 g
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Answer:
Pure water is a non conductor of electricity and dilute acids in their aqueous solutions form free ions, which conducts electricity. Thus when we need to electrolyse water, a dilute acid is added to increase its conductivity.
