Cl-35, as the atomic mass of Chlorine (35.45) is closer to the number 35 than to the number 37. A higher abundance of CL-35 isotope would have caused the atomic number (which is an average of the values of all isotopes of a substances taking relative abundance into consideration) to decrease from 36, which would appear to be the average.
<h3>
Answer:</h3>
52 mm
<h3>
Explanation:</h3>
We are given;
Required to convert it to cm
We are going to use the appropriate conversion factor;
- The units used to measure length include;
Kilometer(km)
10
Hectometer (Hm)
10
Decameter (dkm)
10
Meter(m)
10
Decimeter (dm)
10
Centimeter (cm)
10
Millimeter (mm)
Therefore; the appropriate conversion factor is 10mm/cm
Thus;
5.2 cm will be equivalent to;
= 5.2 cm × 10 mm/cm
= 52 mm
Therefore, the length of magnesium ribbon is 52 mm
Answer:
Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.
Explanation:
-Air becomes increasingly thinner with increasing altitudes.
-As such, oxygen becomes limited at higher altitudes and makes it difficult or almost impossible to breath a condition called hypoxia.
-Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.
Answer:
CH₄
Explanation:
To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.
Step 1: Determine the mass of the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g
Step 2: Calculate the percent by mass of each element
%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%
%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%
Step 3: Divide each percentage by the atomic mass of the element
C: 74.8/12.01 = 6.23
H: 25.2/1.01 = 24.95
Step 4: Divide both numbers by the smallest one, i.e. 6.23
C: 6.23/6.23 = 1
H: 24.95/6.23 ≈ 4
The empirical formula of the hydrocarbon is CH₄.
On Earth it is relatively rare—5.2 ppm by volume in the atmosphere. Most terrestrial helium<span> present today is created by the natural radioactive decay of heavy radioactive elements (thorium and uranium, although there are other examples), as the alpha particles emitted by such decays consist of </span>helium<span>-4 nuclei.</span>
