2.5LX2 M=5 moles
As Molarity X Volume= Total Number of Moles
Valence electrons are shared in covalent bonds
The disease prevention methods are as follows:
- Influenza: Avoid close contact with sick people; Cover your mouth and nose, Clean your hands
- Salmonella: wash hands properly; keep food area clean; cook food properly
- Type 2 Diabetes : exercise regularly; reduce sugar intake; quit smoking
- HIV/AIDS: Avoid sharing of unsterilized sharp objects; avoid unprotected sex; screen blood before transfusion
- Heart disease: avoid trans fat; exercise regularly; avoid smoking
- Whooping cough: wear face masks; wash regularly; get vaccinated
<h3>What is the disease prevention?</h3>
Disease prevention refers to the steps taken to protect oneself from developing a particular disease.
The disease prevention methods for the given diseases is as follows:
- Influenza: Avoid close contact with sick people; Cover your mouth and nose, Clean your hands
- Salmonella: wash hands properly; keep food area clean; cook food properly
- Type 2 Diabetes : exercise regularly; reduce sugar intake; quit smoking
- HIV/AIDS: Avoid sharing of unsterilized sharp objects; avoid unprotected sex; screen blood before transfusion
- Heart disease: avoid trans fat; exercise regularly; avoid smoking
- Whooping cough: wear face masks; wash regularly; get vaccinated
In conclusion, disease prevention is the best policy in health care.
Learn more about disease prevention at: brainly.com/question/13341436
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Heat released from the combustion of 45.5 grams of CH4 in excess oxygen gas is calculated as follows
calculate the number of CH4 moles used = mass/molar mass
= 45.5 grams/ 15 g/mol =2.84 moles
heat released is therefore= moles x 890 kj
= 2.84 x 890=2527.6 kj/mol
Answer:
4
10
Explanation:
The reaction equation is given as;
Ca(OH)₂ → Ca²⁺ + 2OH⁻
Concentration of Ca(OH)₂ = 5 x 10⁻⁵M
Unknown:
pOH of the solution = ?
pH of the solution = ?
Solution:
Solve for the pOH of this solution using the expression below obtained from the ionic product of water;
pOH = ⁻log₁₀[OH⁻]
Ca(OH)₂ → Ca²⁺ + 2OH⁻
1moldm⁻³ 1moldm⁻³ 2 x 1moldm⁻³
5 x 10⁻⁵moldm⁻³ 5 x 10⁻⁵moldm⁻³ 2( 5 x 10⁻⁵moldm⁻³ )
1 x 10⁻⁴moldm⁻³
Therefore;
pOH = -log₁₀ 1 x 10⁻⁴ = 4
Since
pOH + pH = 14
pH = 14 - 4 = 10
