Answer:
1.51367e+10 inches
Explanation:
1 mile = 63360
63360 x 238900 = 15136704000
Hope this helped!
Answer:
84.8 mL
Explanation:
From the question given above, the following data were obtained:
Mass of CuNO₃ = 3.53 g
Molarity of CuNO₃ = 0.330 M
Volume of solution =?
Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:
Mass of CuNO₃ = 3.53 g
Molar mass of CuNO₃ = 63.5 + 14 + (16×3)
= 63.5 + 14 + 48
= 125.5 g/mol
Mole of CuNO₃ =?
Mole = mass / Molar mass
Mole of CuNO₃ = 3.53 / 125.5
Mole of CuNO₃ = 0.028 moles
Next, we shall determine the volume of the solution. This can be obtained as follow:
Molarity of CuNO₃ = 0.330 M
Mole of CuNO₃ = 0.028 moles
Volume of solution =?
Molarity = mole /Volume
0.330 = 0.028 / Volume
Cross multiply
0.330 × Volume = 0.028
Divide both side by 0.330
Volume = 0.028 / 0.330
Volume = 0.0848 L
Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.0848 L = 0.0848 L × 1000 mL / 1 L
0.0848 L = 84.8 mL
Therefore, the volume of the solution is 84.8 mL.
Answer:
mendeleev left a space
Explanation:
so the periodic table can be organize
Answer:
Explanation:
mole of HCl remaining after reaction with CaCO₃
= .3 M of NaOH of 32.47 mL
= .3 x .03247 moles
= .009741 moles
Initial HCl taken = .3 x .005 moles = .0015 moles
Moles of HCl reacted with CaCO₃
= .009741 - .0015 = .008241 moles
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O .
1 mole 2 moles
2 moles of HCl reacts with 1 mole of CaCO₃
.008241 moles of HCl reacts with .5 x .008241 moles of CaCO₃
CaCO₃ reacted with HCl = .5 x .008241 = .00412 moles
the mass (in grams) of calcium carbonate in the tablet
= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )
The molarity of the solution will be 3.45 M
<h3>Further explanation
</h3>
Dilution is the process of adding a solvent to get a more dilute solution.
The moles(n) before and after dilution are the same.
Can be formulated :
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
The initial solution
345 mL of a 2.5 M NaCl solution.
M₁=2.5 M
V₁=345 ml
The solution after boiled :
V₂=250 ml
