Answer:
ACTIVITY 1
Sample 1 has a stronger taste of lemon, and is more sour.
Sample 2 has a sweeter taste, my guess is because there's more sugar:lemon juice ratio.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
Answer:
Samarium
Explanation:
The element Sm describe is called Samarium. This element has unique sets of properties that makes it very unique and distinct.
The lanthanides are found in the f-block on the periodic table of elements.
This element is a moderately hard silvery metal that readily oxidizes in air. It assumes an oxidation state of +3. The element has an atomic number of 62
Answer:
4.549 kg.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 2 x 10⁴ kPa/101.325 = 197.4 atm).
V is the volume of the gas in L (V = 20.0 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 23° C + 273 = 296 K).
<em>∴ n = PV/RT =</em> (197.4 atm)(20.0 L)/(0.0821 L.atm/mol.K)(296 K) = <em>162.5 mol.</em>
- To find the mass of N₂ in the cylinder, we can use the relation:
<em>mass of N₂ = (no. of moles of N₂)*(molar mass of N₂) = </em>(162.5 mol)*(28.0 g/mol) = <em>4549 g = 4.549 kg.</em>
2.
The reason why is because there are 2 available electron spots on the orbital for the oxygen atom. Hydrogen atoms have one proton and one electron, thus, in order to fill the oxygen atom orbital to a full outer shell, a maximum of 2 atoms could bond with the oxygen atoms.
6 electrons (oxygen)+ 1 electron (hydrogen)+ 1 electron (hydrogen)= 8
