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.
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol
HNO3= 1 mol KOH (keep in mind this because it will be used later).
We also know that 0.100 M KOH aqueous
solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the
definition of molarity).
First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/
1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.
Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.
The final answer is </span>(2) 20.0 mL.<span>
Also, this problem can also be done by using
dimensional analysis.
Hope this would help~
</span>
Answer:
7.03 g
Explanation:
Step 1: Write the balanced synthesis reaction
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
Step 2: Calculate the moles corresponding to 32.5 g of N₂
The molar mass of N₂ is 28.01 g/mol.
32.5 g × 1 mol/28.01 g = 1.16 mol
Step 3: Calculate the number of moles of H₂ needed to react with 1.16 moles of N₂
The molar ratio of N₂ to H₂ is 1:3. The moles of H₂ needed are 3/1 × 1.16 mol = 3.48 mol.
Step 4: Calculate the mass corresponding to 3.48 moles of H₂
The molar mass of H₂ is 2.02 g/mol.
3.48 mol × 2.02 g/mol = 7.03 g
Answer:
B. accepted value x 0.1
Explanation:
in the equation provided

Maximum allowed value of percentage error = 10%
put this value in the equation in stead of percentage error we get,

so maximum error = .1 x accepted value
10 % percentage error means the experimental value has 10 % error compared to accepted value.so error will be 10 % of the accepted value
or .1 times of accepted value