Multiply the volume and density together. Multiply your two numbers together, and you'll know the mass of your object. Keep track of the units as you do this, and you'll see that you end up with units of mass (kilograms or grams). Example: We have a diamond with volume 5,000 cm3 and density 3.52 g/cm3
The answer for the following question is option "C".
Option C is not included in the John Dalton's modern theory of an atom.
- "It states atoms of different elements combine to form new compound" but not new elements
Explanation:
According to John's Dalton's modern theory of an atom:
1. All matter is composed of atoms.
2. Atoms cannot be created,destroyed or subdivided in the ordinary chemical reactions.
3. Atoms of one element differ in the properties from atoms of an another element.
(i.e.)Each and every atom of the element has its own unique properties of their own.
4. Atoms of one element combine with the atoms of another element to <u>form new compound.</u>
5. Atoms that make up an element are identical to each other.
Answer:
0.42 g
Explanation:
<u>We have: </u>
pH = 12.10 (25 °C)
V = 800.0 mL = 0.800 L
To find the mass of sodium hydroxide (NaOH) we can use the pH:
![pOH = -log ([OH^{-}])](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%20%28%5BOH%5E%7B-%7D%5D%29%20)
![[OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M](https://tex.z-dn.net/?f=%5BOH%5D%5E%7B-%7D%20%3D%2010%5E%7B-pOH%7D%20%3D%2010%5E%7B-1.90%7D%20%3D%200.013%20M)
Now, we can find the number of moles (η) of OH:
![\eta = ([OH]^{-})*V = 0.013 mol/L * 0.800 L = 1.04 \cdot 10^{-2} moles](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%28%5BOH%5D%5E%7B-%7D%29%2AV%20%3D%200.013%20mol%2FL%20%2A%200.800%20L%20%3D%201.04%20%5Ccdot%2010%5E%7B-2%7D%20moles)
Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.
Finally, with the number of moles we can find the mass of NaOH:
<em>Where M is the molar mass of NaOH = 39.9 g/mol </em>
Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.
I hope it helps you!
