Take note of the whole number beside each substance in the reaction because these will be used in the stoichiometric calculations below:
Molar mass of <span>k[ag(cn)2]:199 g/mol
Molar mass of Zn(Cn)2: 117.44 g/mol
Avogadro's number: 6.022</span>×10²³ molecules/mol
a.)
35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(6.022×10²³ molecules/mol) = <em>5.34×10²² molecules of Zn(Cn)₂</em>
b.)
35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(117.44 g/mol) = <em>10.41 g of Zn(Cn)₂</em>
Answer:
A sample of 830 kg would contain 830000000 mg
Explanation:
Answer:
0.020 moles of
can be formed
Explanation:
1. First determine the number of moles of LiOH.
Molarity is given by the following expression:

Solving for moles of solute:
moles of solute = M * Liters of solution
Converting 175.0mL to L:

Replacing values:
moles of solute = 0.227M*0.175L
moles of solute = 0.040
Therefore there are 0.040 moles of LiOH
2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of
produced:

As the problem says that there are excess of
, the limiting reagent is the LiOH.
can be formed
The reason why the spectrum produced by a 1 gram sample of element z would have the same spectral lines at the same wavelengths as the spectrum produced by a 2gram is because Mass does not determine the spectral lines
hope this helps
Answer:

Explanation:
Hello!
In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

Best regards!