Answer:
Density = mass\ volume
So using this relation we can write volume = mass\density
So volume of uranium = (50.8 \ 18.9 ) cm3
= 2.69 cm3 ( ans)
The right answer to this question is U-235. It is an isotope of Uranium and it is an unsteady heavy metal used in fission reactions because it can run long chains of reactions. I hope this helps.
c. Some health risks are increased by heredity, which manifest under certain environmental conditions.
<h3>The chance of contracting some diseases can arise in particular circumstances.</h3>
The occurrence of certain diseases is influenced by certain environmental factors. Air pollution from a polluted environment increases the chance of allergic people or those with overactive immune systems acquiring respiratory conditions like asthma. A similar genetic flaw in melanin formation increases the risk of skin cancer in people who are exposed to UV rays. As a result, certain circumstances promote the development of disease.
Communities can lessen exposure to diseases and chemicals that have damaging effects on the body by focusing on environmental health. Environmental health treatments can improve everyone's quality of life, but they may be especially noticeable for people with pre-existing health conditions.
Learn more about environmental health here:
brainly.com/question/15119899
#SPJ4
Answer:
1.47 atm
Explanation:
Step 1: Calculate the moles corresponding to 41.6 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
41.6 g × 1 mol/32.00 g = 1.30 mol
Step 2: Convert 30.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 30.0 + 273.15 = 303.2 K
Step 3: Calculate the pressure exerted by the oxygen
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 1.30 mol × (0.0821 atm.L/mol.L) × 303.2 K / 22.0 L = 1.47 atm
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
