Answer:
The uranium in the sample of the compound is radioactive
Explanation:
Some atoms can split on their own. Some split when bombarded by energetic particles. Such atoms are said to be radioactive.
Radioactivity is borne out of the drive of an atom to reach stabillity. Every atom have a specific neutron/proton ratio which ensures stability of the nucleus. A nucleus with a stability ratio different from that which makes it stable will become unstable and split into one or more other nuclei with emissons of energetic particles.
Note: neutrons and protons dictates the mass of an atom. They are located in the nucleus which is the site for nuclear radioactive reactions.
![176.0 \; \text{kJ} \cdot \text{mol}^{-1}](https://tex.z-dn.net/?f=176.0%20%5C%3B%20%5Ctext%7BkJ%7D%20%5Ccdot%20%5Ctext%7Bmol%7D%5E%7B-1%7D)
As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its
can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.
Let the three equations with
given be denoted as (1), (2), (3), and the last equation (4). Let
,
, and
be letters such that
. This relationship shall hold for all chemicals involved.
There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance,
shall resemble the number of
left on the product side when the second equation is directly added to the third. Similarly
Thus
and
![-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%281%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%282%29%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%283%29%3D%20%284%29)
Verify this conclusion against a fourth species involved-
for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.
![a + b = -1/2 + 1/2 = 0](https://tex.z-dn.net/?f=a%20%2B%20b%20%3D%20-1%2F2%20%2B%201%2F2%20%3D%200)
Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.
![\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}](https://tex.z-dn.net/?f=%5CDelta%20H%20_%7B%284%29%7D%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%20%5C%3B%20%5CDelta%20H%20_%7B%281%29%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5C%3B%20%5CDelta%20H%20_%7B%282%29%7D%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5C%3B%20%5CDelta%20H%20_%7B%283%29%7D%5C%5C%5Cphantom%7B%5CDelta%20H%20_%7B%284%29%7D%7D%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%28-628.9%29%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%28-92.2%29%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%28184.7%29%20%5C%5C%5Cphantom%7B%5CDelta%20H%20_%7B%284%29%7D%7D%20%3D%20176.0%20%5C%3B%20%5Ctext%7BkJ%7D%20%5Ccdot%20%5Ctext%7Bmol%7D%5E%7B-1%7D)
150/30 = 5
HF1 20/2 = 10
HF2 10/2 = 5
HF3 5/2 = 2.5
HF4 2.5/2 = 1.25
HF5 1.25/2 = 0.625
Answer: 0.63g
Answer:
![2CH_{4(g)} +3O_{2(g)} -->2CO(g)+4H_{2}O_{(g)}\\CH_{4(g)} +2O_{2(g)} -->CO_{2(g)} +2H_{2}O_{(g)} \\](https://tex.z-dn.net/?f=2CH_%7B4%28g%29%7D%20%2B3O_%7B2%28g%29%7D%20--%3E2CO%28g%29%2B4H_%7B2%7DO_%7B%28g%29%7D%5C%5CCH_%7B4%28g%29%7D%20%2B2O_%7B2%28g%29%7D%20--%3ECO_%7B2%28g%29%7D%20%2B2H_%7B2%7DO_%7B%28g%29%7D%20%5C%5C)
Explanation:
![2CH_{4(g)} +3O_{2(g)} -->2CO(g)+4H_{2}O_{(g)}\\CH_{4(g)} +2O_{2(g)} -->CO_{2(g)} +2H_{2}O_{(g)} \\](https://tex.z-dn.net/?f=2CH_%7B4%28g%29%7D%20%2B3O_%7B2%28g%29%7D%20--%3E2CO%28g%29%2B4H_%7B2%7DO_%7B%28g%29%7D%5C%5CCH_%7B4%28g%29%7D%20%2B2O_%7B2%28g%29%7D%20--%3ECO_%7B2%28g%29%7D%20%2B2H_%7B2%7DO_%7B%28g%29%7D%20%5C%5C)
The incomplete combustion of alkanes, and other fuels actually, happens when there is a limited supply of oxygen. Instead of the fuel burning completely to produce carbon dioxide, it produces carbon monoxide instead.
This gas is harmful to jumans because it combines with haemoglobin in lood and takes up space that belongs to oxygen which can lead to suffocation or even death
The statement that the friend made is not true. most of the mass of the plant is from carbon. the carbon comes from carbon dioxide which is used during photosynthesis. the left over carbon from photosynthesis is used to to help the plant gain mass. there is a process for this which is called cellular respiration