Answer:
A long uniformly charged wire has charge density λ=0.16μλ=0.16μC/m.
Answer:
The molar mass of the liquid 62.89 g/mol
Explanation:
Step 1: Data given
Mass of the sample = 0.1 grams
Temperature = 70°C
Volume = 750 mL
Pressure = 0.05951 atm
Step 2: Calculate the number of moles
p*V = n*R*T
n = (p*V)/(R*T)
⇒ with n = the number of moles gas = TO BE DETERMINED
⇒ with p = The pressure = 0.05951 atm
⇒ with V = The volume of the flask = 750 mL = 0.750 L
⇒ with R = The gasconstant = 0.08206 L*atm/K*mol
⇒with T = the temperature = 70 °C = 343 Kelvin
n = (0.05951 *0.750)/(0.08206*343)
n = 0.00159 moles
Step 3: Calculate molar mass
Molar mass = mass / moles
Molar mass =0.1 gram / 0.00159 moles
Molar mass = 62.89 g/mol
The molar mass of the liquid 62.89 g/mol
Volume ⇒ 50 mL in liters : 50 / 1000 = 0.05 L
Molarity of solution ⇒ 0.15 M
Number of moles:
n = M * V
n = 0.15 * 0.05
n = 0.0075 moles of CuCl2
hope this helps!.
The molar mass of sulfur is 32 g/mol.
The mass of 6.3 moles of Sulfur is 32×6.3=201.6 grams
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Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be 
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio 
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
