Answer: The correct option is B.
Explanation: There are 2 regions of the solar system.
1) Inner region: There are 4 planets which lie in this region: Mercury, Venus, Earth and Mars. The inner region planets are rocky in nature. The orbit time of these planets around the sun is short. They have 0, 1 or 2 satellites in total. There is no ring system in these planets.
2) Outer region: There are 4 planets which lie in this region: Jupiter, Saturn, Uranus and Neptune. The outer region planets are gaseous in nature. The orbit time of these planets around the sun is long. They have usually more number of satellites around them. Ring system in these planets is very common.
Hence, Mars resides in the inner region of the solar system because it has a rocky surface that one could firmly stand on.
Number of atoms : 1.26 x 10²³
<h3>Further explanation </h3>
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
0.21 moles of Al, so n = 0.21
Number of atoms :
The balanced equation of the reaction is:
O3(g) + NO (g) → O2 (g) + NO2 (g)
Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2
If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.
The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.
And the rate of reaction is change in concetration divided by the time.
The change in concentration in O3 is 0.02 M
Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s
The empirical formula is the simplest formula attainable while maintaining the ratio so it will be CH2.
Explanation:
The empirical formula of a chemical compound is the simplistic positive integer ratio of atoms being in a compound. A simple example of this thought is that the empirical formula of sulfur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S2O2.
Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
