The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The partial pressure is
Explanation:
The Partial pressure of
is mathematically represented as

Where
is the total pressure of water with a value of 15.5 mm of Hg
is the partial pressure of water with a value 753 mm of Hg
Now substituting values

The original mass of krypton 81 that is present in the ice is 6.70 grams.
<h3>How do we calculate original mass?</h3>
Original mass of any substance will be calculated as below for the decomposition reaction is:
N = N₀(1/2)ⁿ, where
N = remaining mass of krypton-81 = 1.675g
N₀ = original mass of krypton-81 = ?
n will be calculated as:
n = T/t, where
T = total time period = 458,000 years
t = half life time = 229,000 years
n = 458,000/229,000 = 2
Now putting all these values on the above equation, we get
N₀ = 1.675 / (1/2)²
N₀ = 6.70 g
Hence required mass is 6.70 g.
To know more about half life time, visit the below link:
brainly.com/question/2320811
Answer:
a) IUPAC Names:
1) (<em>trans</em>)-but-2-ene
2) (<em>cis</em>)-but-2-ene
3) but-1-ene
b) Balance Equation:
C₄H₁₀O + H₃PO₄ → C₄H₈ + H₂O + H₃PO₄
As H₃PO₄ is catalyst and remains unchanged so we can also write as,
C₄H₁₀O → C₄H₈ + H₂O
c) Rule:
When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.
d) C is not Geometrical Isomer:
For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.