Answer:
Mass = 1274 .64 g it would be option C if it is converted into kilogram
1274 .64 / 1000 = 1.27 Kg
Explanation:
Given data:
Number of moles of C₂₀H₄₂ = 4.52 mol
Molar mass of carbon = 12 g/mol
Molar mass of hydrogen = 1.0 g/mol
Mass of C₂₀H₄₂ = ?
Solution:
Number of moles = mass / molar mass
Molar mass = 20× 12 + 42× 1.0 = 282 g/mol
Now we will put the values in formula:
Number of moles = mass / molar mass
4.52 mol = mass / 282 g /mol
Mass = 4.52 mol × 282 g/mol
Mass = 1274 .64 g
Answer:
[NaCH₃COO] = 2.26M
Explanation:
17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)
Let's determine the volume of solution, by density
Mass of solution / Volume of solution = Solution density
100 g / Volume of solution = 1.09 g/mL
100 g / 1.09 g/mL = 91.7 mL
17 grams of solute is contained in 91.7 mL
Molarity (M) = Mol of solute /L of solution
91.7 mL / 1000 = 0.0917L
17 g / 82 g/m = 0.207 moles
Molariy = 0.207 moles / 0.0917L → 2.26M
1) <span>NaNO3 and H2O - no reaction , it is dissolution
2) no hydrogen to make water
3) </span><span>Fe(OH)3 (base) and H2SO4(acid))
base +acid ----> salt +water
4) </span><span>Li2O and Ba(OH)2
basic oxide and base ----> no reaction
so Answer number 3)
</span> 2Fe(OH)3 +3 H2SO4 ------> Fe2(SO4)3 + 6H2O<span>
</span>
Answer:
It is either the third or fourth statement.
Explanation:
This is because exothermic reactions give off heat.
Answer:
the change in energy of the gas mixture during the reaction is 227Kj
Explanation:
THIS IS THE COMPLETE QUESTION BELOW
Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -92kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.
From thermodynamics
ΔE= q + w
Where w= workdone on the system or by the system
q= heat added or remove
ΔE= change in the internal energy
q=+ 319kJ ( absorbed heat is + ve
w= -92kJ
If we substitute the given values,
ΔE= 319 + (-92)= 227 Kj
With the increase in enthalpy and there is absorbed heat, hence the reaction is an endothermic reaction.
