<span>Methink this is a neutralization reaction where an acid reacts with a base to form salt and water; but if we must balance the equation, we need to know what the product would be?
So our base LiOH reacts with our acid Tetraoxosulphate (VI) H2SO4. The reaction produces salt and water as evidence. LiOH + H2SO4 gives Li2SO4 and H20. We need to make sure that the total atoms on the LHS and RHS balanced. So adding two moles of LiOH, we have 2LiOH + H2SO4 produces LiSO4 + H20. The eqn isn't balanced yet as there's 2 moles of deficit H2 atoms on the RHS,
So our final reaction we have 2LiOH + H2SO4 gives Li2SO4 + 2H2O. Hence our answer is C</span>
Because in heat rises faster on the floor than on the ceiling or side of a wall.
Explanation:
यह लेख एक आधार है। जानकारी जोड़कर इसे बढ़ाने में विकिपीडिया की मदद करें।
उस पदार्थ या यौगिक को अभिकर्मक (reagent) कहते हैं जो किसी तंत्र में रासायनिक अभिक्रिया उत्पन्न करने के लिये डाला या मिलाया जाता है। उस पदार्थ को भी अभिकर्मक कहेंगे जिसे यह जांचने के लिये मिलाया जाता है कि कोई अभिक्रिया होती है या नहीं। इस तरह के कुछ वैश्लेषिक अभिकर्मक हैं - फेहलिंग का अभिकर्मक (Fehling's reagent), मिलॉन का अभिकर्मक (Millon's reagent) तथा टॉलीन का अभिकर्मक (Tollens' reagent)।
Magnesium would be more reactive.
Answer: Boron is the element which has properties of both metals and nonmetals.
Explanation:
Metals are defined as the elements which loose electrons to attain stable electronic configuration. They attain positive charge and form cation. Example: Zinc (Zn), Aluminium (Al) , copper (Cu)
Non-metals are defined as the elements which gain electrons to attain stable electronic configuration. They attain negative charge and form anion. Example: Chlorine (Cl) , Sulphur (S)
Metalloids are defined as the elements which show properties of both metals and non-metals. There are 7 metalloids in the periodic table. They are Boron (B) , Silicon (Si) , Germanium (Ge) , Arsenic (As) , Antimony (Sb), Tellurium (Te) and Polonium (Po).
Thus boron is the element which has properties of both metals and nonmetals.
