Answer:
Density = 8.75ml
Explanation:
Density = Mass / Volume
In this problem ...
Mass = 28 grams
Volume = 31.4ml - 28.2ml = 32ml (water displacement also)
∴Density = mass /volume = 28g/32ml = 8.75g/ml
His the answer I hope it helps
Answer:
The answer to your question is 160 g of Fe₂O₃
Explanation:
Data
mass of Fe = 112 g
mass of CO = in excess
mass of Fe₂O₃ = ?
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
Process
1.- Calculate the molar mass of Fe₂O₃ and Fe
Molar mass Fe₂O₃ = (56 x 2) + (16 x 3) = 112 + 48 = 160 g
atomic mass of Fe = 56
2.- Use proportions to calculate the mass of Fe₂O₃ needed
160 g of Fe₂O₃ ------------------- 2(56) g of Fe
x g of Fe₂O₃ ------------------ 112 g of Fe
x = (112 x 160) / 2(56)
x = 17920/112
x = 160 g of Fe₂O₃
Most atoms have more neutrons than protons which accounts for basically half of the nucleuses mass.
First, we have to remember the molarity formula:
![M=\text{ }\frac{moles\text{ of solute}}{L\text{ solution}}](https://tex.z-dn.net/?f=M%3D%5Ctext%7B%20%7D%5Cfrac%7Bmoles%5Ctext%7B%20of%20solute%7D%7D%7BL%5Ctext%7B%20solution%7D%7D)
Part 1:
In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:
![M.W_{NaNO_3}=\text{ 23+14+16*3= 85 g/mol}](https://tex.z-dn.net/?f=M.W_%7BNaNO_3%7D%3D%5Ctext%7B%2023%2B14%2B16%2A3%3D%2085%20g%2Fmol%7D)
Then, we calculate the moles present in the solution:
![3.976\text{ g NaNO}_3\text{ * }\frac{1\text{ mol}}{85\text{ g}}=\text{ 0.04678 mol NaNO}_3](https://tex.z-dn.net/?f=3.976%5Ctext%7B%20g%20NaNO%7D_3%5Ctext%7B%20%2A%20%7D%5Cfrac%7B1%5Ctext%7B%20mol%7D%7D%7B85%5Ctext%7B%20g%7D%7D%3D%5Ctext%7B%200.04678%20mol%20NaNO%7D_3)
Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):
![M=\frac{0.04678\text{ mol}}{200\text{ mL*}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.2339 M}](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B0.04678%5Ctext%7B%20mol%7D%7D%7B200%5Ctext%7B%20mL%2A%7D%5Cfrac%7B1%5Ctext%7B%20L%7D%7D%7B1000%5Ctext%7B%20mL%7D%7D%7D%3D%5Ctext%7B%200.2339%20M%7D)
The molarity of this solution equals 0.2339 M.
Part 2:
In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:
![M=\text{ }\frac{0.04678\text{ mol}}{275\text{ mL *}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.1701 M}](https://tex.z-dn.net/?f=M%3D%5Ctext%7B%20%7D%5Cfrac%7B0.04678%5Ctext%7B%20mol%7D%7D%7B275%5Ctext%7B%20mL%20%2A%7D%5Cfrac%7B1%5Ctext%7B%20L%7D%7D%7B1000%5Ctext%7B%20mL%7D%7D%7D%3D%5Ctext%7B%200.1701%20M%7D)
So, the molarity of this solution is 0.1701 M.