Answer:
The velocity of the particle is 2 m/s,
Explanation:
Kinetic energy is defined as energy of the body due to its motion. It is given by :
Where :
m = mass of the object
v = velocity of the object
We have , particle with mass m and its kinetic energy is twice its mass.
And unit of velocity are m/s , so the velocity of the particle is 2 m/s.
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
Answer:
5.63 mol.
Explanation:
- The balanced chemical equation between NO₂ and H₂O is:
<em>3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g),
</em>
It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.
<em>Water is present as an excess reactant and NO₂ is limiting reactant.</em>
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- To find the no. of moles of HNO₃ produced:
3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.
8.44 mol of NO₂ produces → ??? mol of HNO₃.
∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.
Answer:
The answer to your question is: Excess oxygen = 2.3 mol
Explanation:
Data
ZnS = 5 mol
O2 = 9.8 mol
Excess reactant = ?
Balanced reaction
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
MW ZnS = 65 + 32 = 97 x 2 = 194 g
MW O2 = 16 x 6 = 96 g
2 mol of ZnS ------------------- 3 mol O2
Ratio from the reaction = 3 mol O2/ 2 mol ZnS
= 1.5
Ratio from the quantities in the experiment = 9.8 mol O2 / 5 mol of ZnS
= 1.96
Excess reactant = Oxygen because the ratio increases
2 mol of ZnS ------------------- 3 mol O2
5 mol of ZnS ------------------- x
x = (5 x 3) / 2
x = 7.5 mol of O2
Excess Oxygen = 9.8 mol - 7.5 mol
Excess oxygen = 2.3 mol
Answer:
There will be no observed impact of adding twice as much Na2CO3 on the product
Explanation:
Stoichiometry gives the relationship between reactants and products in terms of mass, mole and volume.
If we consider the stoichiometry of the reaction, we will discover that the reaction occurs in a 1:1 ratio. This implies that use of twice the amount of Na2CO3 will only lead to an excess of Na2CO3 making the other reactant the limiting reactant. Once the other reactant is used up, the reaction quenches.
Hence, use of twice as much Na2CO3 has no impact on the quantity of product produced.
