Answer:
760 mmHg
Explanation:
Step 1: Given data
- Partial pressure of nitrogen (pN₂): 592 mmHg
- Partial pressure of oxygen (pO₂): 160 mmHg
- Partial pressure of argon (pAr): 7 mmHg
- Partial pressure of the trace gas (pt): 1 mmHg
Step 2: Calculate the atmospheric pressure
Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.
P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg
The red colour is the limiting reactant.
Red-blue colour ball and two white balls attached together are reactants.
Red-blue colour ball and two white and one red colour ball attached to each other are products.
<h3>What is a limiting reagent?</h3>
The reactant that is entirely used up in a reaction is called a limiting reagent.
A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products.
A product is a substance that is present at the end of a chemical reaction.
Hence,
The red colour is the limiting reactant.
Red-blue colour ball and two white balls attached together are reactants.
Red-blue colour ball and two white and one red colour ball attached to each other are products.
Learn more about limiting reagents here:
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Answer:
Explanation: Mendeleev arranged the elements on the basis of their atomic mass. Melting and boiling point were used as the physical characteristics in deciding the position of elements. He arranged the elements and wrote the formula of their oxides and hydrides which seemed to possess same chemical formula.
Explanation:
Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3
The answer would be b) GAS