Molar mass of Cu(NO3)2 = 187.56 g/mol
Mass = 122 g
number of moles:
n = m / mm
n = 122 / 187.56
n = 0.650 moles
n * 6.02x10²³ =
0.650 * 6.02x10²³ = 3.913 x 10²³ molecules
hope this helps!.
Answer:
∆S = ∆H / T = 17.2 kJ/mol (298 K)
∆S = 0.0577 kJ/mol K
∆S = 0.0577 kJ/mol K x 1.41 moles = 0.0814 kJ/K = 81.4 J/K
1.41 mol x 17.2 kJ/mol = 24.25 kJ
∆S = 24.25 kJ / 298 K = 0.0814 kJ/K = 81.4 J/K
Explanation:
Answer:
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Explanation:
<u>1. Balanced molecular equation</u>
![2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O](https://tex.z-dn.net/?f=2HNO_3%2BBa%28OH%29_2%5Crightarrow%20Ba%28NO_3%29_2%2B2H_2O)
<u>2. Mole ratio</u>
![\dfrac{2molHNO_3}{1molBa(OH)_2}](https://tex.z-dn.net/?f=%5Cdfrac%7B2molHNO_3%7D%7B1molBa%28OH%29_2%7D)
<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:
![\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28](https://tex.z-dn.net/?f=%5Cdfrac%7B0.0600molHNO_3%7D%7B0.0217molBa%28OH%29_2%7D%5Capprox0.28)
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
Answer:
1. decomposition
2. combustion
3.single replacement
4. combination
5. double replacement
Explanation:
1. one compound is split into 2 elements
2. co2 and h20 was the product of the reaction
3. cu is replaced with co
4. 2 compounds become one compound
5. ca is replaced with na and na is replaced with ca