Answer;
294.13 amu
Solution;
-293nv is 293.10 amu and that of 295nv is 295.45 amu
293.05+295.2=588.25
588.25/2= 294.13 amu
an amu of 294.13 using significant figures
Answer:
.081 g of O2
Explanation:
4Cr + 3O2 -----> 2Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2
The question is incomplete, the complete question is;
Chlorine monoxide accumulates in the stratosphere above Antarctica each winter 3nd plays a key role the formation of the ozone hole above the South Pole each spring Eventually. CIO decomposes acco to the equation: 2CIO(g) rightarrow CL2(g) + O2(g) The second-order rate constant for the decomposition of CIO is 6950000000 M-1 s-1 at a particular temperature Determine the half-life of CIO when its initial concentration is .0000000185 M
Answer:
7.8 * 10^-3 s
Explanation:
Given that the half life of a second order reaction is obtained from the formula;
t1/2 = k-1[A]o-1
t1/2 = 1/k[A]o
second order rate constant (k) = 6950000000 M-1 s-1
initial concentration ([A]o) =0.0000000185 M
t1/2 = 1/6950000000 * 0.0000000185
t1/2 = 7.8 * 10^-3 s
Answer: d
Explanation: For an example, many scientists use scientific notation when referring to the sun and moon. They use the power of ten to refer to the large numbers of miles away from the earth.
Answer: Left Arrow Far right arrow and center left arrow
Explanation:
These three arrows are correct due to heat transfers to cooler temp environments which allows the expansion of speed for molecules in their environment.
