Answer:
Therefore the density of the sheet of iridium is 22.73 g/cm³.
Explanation:
Given, the dimension of the sheet is 3.12 cm by 5.21 cm.
Mass: The mass of an object can't change with respect to position.
The S.I unit of mass is Kg.
Weight of an object is product of mass of the object and the gravity of that place.
Density: The density of an object is the ratio of mass of the object and volume of the object.
[S.I unit of mass= Kg and S.I unit of m³]
Therefore the S.I unit of density = Kg/m³
Therefore the C.G.S unit of density=g/cm³
The area of the sheet is = length × breadth
=(3.12×5.21) cm²
=16.2552 cm²
Again given that the thickness of the sheet is 2.360 mm =0.2360 cm
Therefore the volume of the sheet is =(16.2552 cm²×0.2360 cm)
=3.8362272 cm³
Given that the mass of the sheet of iridium is 87.2 g.
=22.73 g/cm³
Therefore the density of the sheet of iridium is 22.73 g/cm³.
It is important to ensure that treated water remains safe to drink because water does not last forever as it can gain bacteria and organisms in it. To make sure storage of water is safe is to simply add chlorine again over a period of time.
-never store in direct sunlight
-containment of the water is clean
-make sure chemicals or anything that can contaminate it doesn't come near it
Answer:
<u>1. Net ionic equation:</u>
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s)
<u />
<u>2. Volume of 1.0M AgNO₃</u>
Explanation:
1. Net ionic equation for the reaction of NaCl with AgNO₃.
i) Molecular equation:
It is important to show the phases:
- (aq) for ions in aqueous solution
- (s) for solid compounds or elements
- (g) for gaseous compounds or elements
- NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)
ii) Dissociation reactions:
Determine the ions formed:
- NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
- AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
- NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)
iii) Total ionic equation:
Substitute the aqueous compounds with the ions determined above:
- Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
iv) Net ionic equation
Remove the spectator ions:
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer
2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl
i) Determine the number of moles of AgNO₃
The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl
The number of moles of AgCl is determined using the molar mass:
- number of moles = mass in grams / molar mass
- molar mass of AgCl = 143.32g/mol
- number of moles = 5.84g / (143.32g/mol) = 0.040748 mol
ii) Determine the volume of AgNO₃
- molarity = number of moles of solute / volume of solution in liters
- V = 0.040748mol / (1.0M) = 0.040748 liter
- V = 0.040748liter × 1,000ml / liter = 40.748 ml
Round to two significant figures: 41ml ← answer
I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O.
</span>
For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is <span> 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
25g/ (73g/mol) * (87g/mol) * 1/1= 29.8 grams</span>
