To accurately measure the final volume of a solution in a burette, it is important that you read of the measurement in the lower meniscus like for water and for substances like mercury you read it on the upper meniscus. This is the standard way of reading volumes in laboratory apparatus. Also, it should be that the apparatus standing vertically in a plain surface.
<u>Answer:</u> The net ionic equation is written below.
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of ammonium carbonate and lead nitrate is given as:
Ionic form of the above equation follows:
As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation is written above.
Answer:
<h2>0.52 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
volume = final volume of water - initial volume of water
volume = 35 - 8 = 27 mL
We have
We have the final answer as
<h3>0.52 g/mL</h3>
Hope this helps you
Answer:
m = 700 g
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
Density of octane = 0.700 g/cm³
Volume = 1 L
Mass = ?
Formula:
D=m/v
D= density
m=mass
V=volume
First of all we will convert the volume in cm³ because density is given in g/cm³ unit.
1 L = 1000 cm³
Now we will put the values in formula:
d= m/v
m = v × d
m = 1000 cm³ × 0.700 g/cm³
m = 700 g
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
