Answer: i’m pretty sure it’s 560 sorry if I'm wrong but can I get brainlest
Explanation:
Answer:
0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment
Explanation:
The concentration of the solution is given by :
![[C]=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}](https://tex.z-dn.net/?f=%5BC%5D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20compound%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20in%20Liters%7D%7D)
We have:
Concentration of copper (II) nitrate solution = ![[Cu(NO_3)_2]=2.41 M](https://tex.z-dn.net/?f=%5BCu%28NO_3%29_2%5D%3D2.41%20M)
The volume of solution = 4.12 mL
1 mL= 0.001 L
![4.12 mL= 4.12\times 0.001 L= 0.00412 L](https://tex.z-dn.net/?f=4.12%20mL%3D%204.12%5Ctimes%200.001%20L%3D%200.00412%20L)
Moles of copper (II) nitrate in solution = n
![2.41=\frac{n}{0.00412 L}=0.0099292 mol](https://tex.z-dn.net/?f=2.41%3D%5Cfrac%7Bn%7D%7B0.00412%20L%7D%3D0.0099292%20mol)
Moles of copper (II) nitrate in solution = 0.0099292 mol
1 Mole of copper(II) nitrate has 1 mole of copper then 0.0099292 moles of copper(II) nitrate will have :
![1\times 0.0099292 mol= 0.0099292 \text{ mol of Cu}](https://tex.z-dn.net/?f=1%5Ctimes%200.0099292%20mol%3D%200.0099292%20%5Ctext%7B%20mol%20of%20Cu%7D)
Mass of 0.0099292 moles of copper:
![=0.0099292 mol\times 63.55 g/mol=0.63100 g\approx 0.631 g](https://tex.z-dn.net/?f=%3D0.0099292%20mol%5Ctimes%2063.55%20g%2Fmol%3D0.63100%20g%5Capprox%200.631%20g)
This mass of copper present in the solution is the theoretical mass of copper present in the given copper(II) nitrate solution.
0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment
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The formula for molality---> m = moles solute/ Kg of solvent
the solute here is NH₃ because it's the one with less amount. which makes water the solvent.
1) let's convert the grams of NH₃ to moles using the molar mass
molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol
15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃
2) let's convert the grams of water into kilograms (just divide by 1000)
250.0 g= 0.2500 kg
3) let's plug in the values into the molality formula
molality= mol/ Kg---> 0.881 mol/ 0.2500 kg= 3.52 m
bromine
Explanation:
halogens are a group of elemnts simlar to eachother
flourine, chlorine, and bromine