How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g.

Question: How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g. Use correct number of significant digits;

Solution: Mol weight of Ca(NO3)2 = 90 + 28 +96
= 164 g/mol
100 g of Ca(NO3)2 = 0.6097 mol of Ca(NO3)2
1 mol of Ca(NO3)2 contains 6 number of oxygen atoms.
so, 0.6097 moles of Ca(NO3)2 contains = 0.6097 * 6 = 3.6582 Now, mole of each ‘O’ atom is 16 g/mol So, 100 g of Ca(NO3)2 contains = 3.6582*16 = 58.5312 g of oxygen.

A superconductor is a substance that is able to conduct electricity without resistance

Question: A superconductor is a substance that is able to conduct electricity without resistance, a property that is very desirable in the construction of large electromagnets. Metals have this property if cooled to temperatures a few degrees above absolute zero, but this requires the use of expensive liquid helium (boiling point 4 K). Scientists have discovered materials that become superconductors at higher temperatures, but they are ceramics. Their brittle nature has so far prevented them from being made into long wires. A recently discovered compound of magnesium and boron, which consists of 52.9% Mg and 47.1% B, shows special promise as a high-temperature superconductor because it is inexpensive to make and can be fabricated into wire relatively easily. What is the formula of this compound?

Solution:

Let us take 100 g of compound
Then, mass of Mg = 52.9g
and mass of boron = 47.1g
Moles of Mg = Mass/Molar Mass = 52.9g/24.31g/mol
= 2.18 mol
and moles of B = 47.1g/10.81g/mol
= 4.36 mol
So, the molar ratio of Mg and B:
nMg/nB = 2.18 mol/4.36 mol = 1/2
Formula of the compound is MgB2

A 1200-kg two-wheel-drive car drives up θ-15° incline

Question: A 1200-kg two-wheel-drive car drives up θ-15° incline. If the coefficient of static friction between the tires and the ground is 0.7, find the maximum acceleration of the car as well as the normal forces on the pairs of tires at R and F in the separate cases where the car is a) rear-wheel drive and b) front-wheel drive. Neglect any rolling resistance and the mass of the wheels 0.9 m’ 1.6 m 0.76 m

Solution: