Answer:
Q.1
Given-
Volume of solution-1 L
Molarity of solution -6M
to find gms of AgNO3-?
Molarity = number of moles of solute/volume of solution in litre
number of moles of solute = 6×1= 6moles
one moles of AgNO3 weighs 169.87 g
so mass of 6 moles of AgNO3 = 169.87×6=1019.22
so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution
Answer:A colloid is a mixture in which one substance consisting of microscopically dispersed insoluble particles is suspended throughout another substance.
Explanation:
And yes I would love to talk
Answer:
Natural resources are not evenly distributed all over the world. Some places are more endowed that others — for instance, some regions have lots of water (and access to ocean and seas). Others have lots of minerals and forestlands. Others have metallic rocks, wildlife, fossil fuels and so on.
Explanation:
The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,
Mg + 2HCl --> H2 + MgCl2
The number of moles of HCl that is needed for the reaction is calculated below.
n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
n = 0.0385 mols HCl
From the given concentration, we calculate for the required volume.
V = 0.0385 mols HCl/(0.650 mols/L)
V = 0.05926 L or 59.26 mL
<em>Answer: 59.26 mL of HCl</em>
The amount of oxygen that are produced when 1.06 grams of potassium chlorate decompose completely is 0.64 grams.
<h3>What is the relation between mass & moles?</h3>
Relation between the mass and moles of any substance will be represented as:
- n = W/M, where
- W = given mass
- M = molar mass
Moles of potassium chlorate = 1.66g / 122.5g/mol = 0.0135mole
Given chemical reaction is:
2KClO₃ → 2KCl + 3O₂
From the stoichiometry of the reaction, it is clear that:
2 moles of KClO₃ = produces 3 moles of O₂
0.0135 moles of KClO₃ = produces (3/2)(0.0135)=0.02 moles of O₂
Mass of oxygen = (0.02mol)(32g/mol) = 0.64 g
Hence produced mass of oxygen is 0.64 grams.
To now more about mass & moles, visit the below link:
brainly.com/question/18983376
#SPJ1
