The question is: 10 g of carbonic acid H2CO3 are dissolved in 150 g of water. Determine the% m / m concentration of that solution?
Answer: The% m / m concentration of that solution is 6.66%.
Explanation:
Given: Mass of solute = 10 g
Mass of solvent = 150 g
Formula used to calculate the %m/m is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the% m / m concentration of that solution is 6.66%.
Answer:
Explanation:
I want to believe that your question is how to find an 8M carbon chloride solution using the given mass of the compound.
The problem clearly identifies with concentration of a substance. Molarity is one of the ways of expressing concentration. It is the amount of solute of a substance per volume of solution.
Molarity =
To find the right mix that will give us 8M carbon chloride solution, let's explore how to figure out the appropriate volume.
Clearly given is the mass of the compound, convert to number of moles.
Number of moles =
Molar mass of carbon chloride(CCl₄) = 12 + 4(35.5) = 154g/mol
Number of moles = = 0.57moles
To find the volume;
8 =
volume = 0.071L or 71mL
The mass of a lead ball with a radius of 0.50mm is 5.9714226× g
The density of a substance is defined as the mass of substance per unit of its volume. Density is represented by the symbol, ρ.
The formula for density is -
Density (ρ) = <u> mass (m) </u>
volume (v)
Given :
Density of lead = 11.4 g/cm³
Radius of lead ball = 0.50 mm = 0.05 cm
Since,
Vsphere = 4/3 π r³
v = <u>4 </u>× 3.14 × (0.05)³
3
v = 4.18 x (1.25 x 10⁻³)
v = 5.225 x 10⁻³
Since,
Density = <u> mass </u>
volume
11.4 g/cm³ = <u> mass </u>
5.225 x 10⁻³
mass = 11.4 x (5.225 x 10⁻³)
mass = 5.9565 x 10⁻² or 0.05955 g
So, the mass of a lead ball with a radius of 0.50mm is 5.9565 x 10⁻² g
To learn more about density,
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Answer:
a) 0
Explanation:
Each of the small dots surrounding the C1 represents one electron. These are where electrivity comes from. Since there is the same number of electrons in both atoms, the difference is 0 (because 6 electrons-6electrons= 0).