Answer:
CuSO4 + 2 NaOH = Cu(OH)2 + Na2SO4
<h3><u>Answer</u>;</h3>
Actual yield = 46.44 g
<h3><u>Explanation;</u></h3>
1 mole of water = 18 g/mol
Therefore;
The experimental yield = 2.58 moles
equivalent to ; 2.58 × 18 = 46.44 g
The theoretical value is 47 g
Percentage yield = 46.44/47 × 100%
= 98.8%
The questions asks for actual yield = 46.44 g
Answer:
The age of the sample is 4224 years.
Explanation:
Let the age of the sample be t years old.
Initial mass percentage of carbon-14 in an artifact = 100%
Initial mass of carbon-14 in an artifact = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final mass percentage of carbon-14 in an artifact t years = 60%
Final mass of carbon-14 in an artifact = ![[A]=0.06[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D0.06%5BA_o%5D)
Half life of the carbon-14 = 

![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%5Ctimes%20t%7D)
![0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}](https://tex.z-dn.net/?f=0.60%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7B5730%20year%7D%5Ctimes%20t%7D)
Solving for t:
t = 4223.71 years ≈ 4224 years
The age of the sample is 4224 years.
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O
Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...
Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...
Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....
Do the calculation
Answer:
11.66 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If P and T are constant, and have different values of n and V:
<em>(V₁n₂) = (V₂n₁).</em>
V₁ = 25.5 L, n₁ = 3.5 mol.
V₂ = ??? L, n₂ = 3.5 mol - 1.9 mol = 1.6 mol.
<em>∴ V₂ = (V₁n₂)/(n₁)</em> = (25.5 L)(1.6 mol)/(3.5 mol) =<em> 11.66 L.</em>