The answer is B. The less mass then the less the gravity pulls on it
Answer:b
Explanation: it’s in the picture just go counterclockwise
Answer:
1. 3.70 g Na₂CO₃·10H₂O
2. 50.0 mL of the first solution
Explanation:
1. Prepare the solution
(a) Calculate the molar mass of Na₂CO₃·10H₂O
![\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrr%7D%5Ctextbf%7BAtoms%7D%20%26%5Ctextbf%7BM%7D_%7B%5Ctextbf%7Br%7D%7D%20%26%20%5Ctextbf%7BMass%2Fu%7D%5C%5C%5Ctext%7B2Na%7D%20%26%202%5Ctimes22.99%20%26%2045.98%5C%5C%5Ctext%7B1C%7D%20%26%201%5Ctimes%2012.01%20%26%2012.01%5C%5C%5Ctext%7B13O%7D%2613%20%5Ctimes16.00%20%26%20208.00%5C%5C%5Ctext%7B20H%7D%2620%20%5Ctimes%201.008%20%26%2020.16%5C%5C%26%5Ctext%7BTOTAL%20%3D%7D%20%26%20%5Cmathbf%7B286.15%7D%5C%5C%5Cend%7Barray%7D)
The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.
(b) Calculate the moles of Na₂CO₃·10H₂O
![\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%7D%5C%5C%3D%20%5Ctext%7B0.250%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B0.0500%20mol%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%5C%5C%5C%5C%3D%20%5Ctext%7B0.0125%20mol%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%7D)
(c) Calculate the mass of Na₂CO₃·10H₂O
![\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%20%7D%5C%5C%3D%20%5Ctext%7B0.012%2050%20mol%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%20%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B296.15%20g%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%7D%7D%7B%5Ctext%7B1%20mol%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%7D%7D%5C%5C%5C%5C%3D%20%5Ctext%7B3.70%20g%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%7D%5C%5C%5Ctext%7BYou%20need%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B3.70%20g%7D%7D%24%20of%20Na%24_%7B2%7D%24CO%24_%7B3%7D%5Ccdot%2410H%24_%7B2%7D%24O%7D)
2. Dilute the solution
We can use the dilution formula to calculate the volume needed.
V₁c₁ = V₂c₂
Data:
V₁ = ?; c₁ = 0.0500 mol·L⁻¹
V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹
Calculation:
![\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DV_%7B1%7Dc_%7B1%7D%20%26%20%3D%20%26%20V_%7B2%7Dc_%7B2%7D%5C%5CV_%7B1%7D%5Ctimes%20%5Ctext%7B0.0500%20mol%2FL%7D%20%26%20%3D%20%26%20%5Ctext%7B100%20mL%7D%20%5Ctimes%5Ctext%7B0.0250%20mol%2FL%7D%5C%5C0.0500V_%7B1%7D%26%20%3D%20%26%20%5Ctext%7B2.500%20mL%7D%5C%5CV_%7B1%7D%26%3D%26%20%5Cdfrac%7B%5Ctext%7B2.500%20mL%7D%7D%7B0.0500%7D%5C%5C%5C%5C%26%20%3D%20%26%20%20%5Ctext%7B50.0%20mL%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BYou%20need%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B50.0%20mL%7D%7D%24%20of%20the%20concentrated%20solution.%7D)
D. Salt and water
Explanation:
Acid + Alkali -> Salt + water