Answer:
B. To Identify the half reactions for the equation
The number of calories that are required to change the temperature of 2.18 g of water from 15.3 c to 69.5 c is <u>118.16 cal</u>
<u><em> calculation</em></u>
- Heat in calories = MCΔ T where,
- M(mass)= 2.18 g
- C(specific heat capacity)= 1.00 cal/g/c
- ΔT( change in temperature)= 69.5- 15.3 =54.2 c
heat is therefore= 2.18 g x 1.00 cal/g/c x 54.2 c=118.16 cal
Answer:
37.25 grams/L.
Explanation:
- Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.
<em>M = (no. of moles of KCl)/(volume of the solution (L))</em>
<em></em>
∵ no. of moles of KCl = (mass of KCl)/(molar mass of KCl)
∴ M = [(mass of KCl)/(molar mass of KCl)]/(volume of the solution (L))
∴ (mass of KCl)/(volume of the solution (L)) = (M)*(molar mass of KCl) = (0.5 M)*(74.5 g/mol) = 37.25 g/L.
<em>So, the grams/L of KCl = 37.25 grams/L.</em>
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.
Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
