<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
The characteristics of the α and β particles allow to find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.
The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.
The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.
Thorium has several isotopes, with different rates and types of emission:
- ²³²Th emits α particles, it is the most abundant 99.9%
- ²³⁴Th emits β particles, exists in small traces.
In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.
Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.
In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
Learn more about radioactive emission here: brainly.com/question/15176980
The volume (in cm³) gained by a person who gains 11.8 lb of fat is 5830.49 cm³
<h3>What is density? </h3>
The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as
Density = mass / volume
<h3>How to convert pounds to grams </h3>
1 lb = 453.592 g
Therefore,
11.8 lb = 11.8 × 453.592
11.8 lb = 5352.3856 g
<h3>How to determine the volume </h3>
- Mass = 5352.3856 g
- Density = 0.918 g/cm³
- Volume =?
Volume = mass / density
Volume = 5352.3856 / 0.918
Volume = 5830.49 cm³
Learn more about density:
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