the answer is d hope this helps
Capture all of the smoke and weight it. it will weigh exactly the same before and after you burn it but will just be CO2 and H2O gas.
The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
Answer:
Mass = 2.77 g
Explanation:
Given data:
Mass of HCl = 2 g
Mass of CaCl₂ produced = ?
Solution:
Chemical equation:
2HCl + Ca → CaCl₂ + H₂
Number of moles of HCl:
Number of moles = mass / molar mass
Number of moles = 2 g/ 36.5 g/mol
Number of moles = 0.05 mol
now we will compare the moles of HCl with CaCl₂.
HCl : CaCl₂
2 : 1
0.05 : 1/2×0.05 = 0.025 mol
Mass of CaCl₂:
Mass = number of moles × molar mass
Mass = 0.025 mol × 110.98 g/mol
Mass = 2.77 g
Answer:
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Explanation:
<u>1. Balanced molecular equation</u>
<u>2. Mole ratio</u>
<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:
Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
