Answer: Mass of 
produced in this reaction was 6.56 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Mass or reactants = Mass of 
+ mass of 
= 16.00 + 64.80 = 80.80 g
Mass of products = mass of aqueous solution + mass of + = 74.24 + x g
Mass or reactants = Mass of products
80.80 g = 74.24 + x g
x = 6.56 g
Thus mass of produced in this reaction was 6.56 grams
Answer:
After complete reaction, 0.280 moles of ammonia are produced
Explanation:
Step 1: Data given
Number of moles N2 = 0.140 moles
Number of moles H2 = 0.434 moles
Step 2: The balanced equation
N2(g) + 3H2 (g) ⟶ 2NH3 (g)
Step 3: Calculate the limiting reactant
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
N2 is the limiting reactant. It will completely be consumed (0.140 moles).
H2 is in excess. There will react 3*0.140 = 0.420 moles
There will remain 0.434 - 0.420 = 0.014 moles
Step 4: Calculate moles NH3
For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3
After complete reaction, 0.280 moles of ammonia are produced
Answer:
Starch is the stored form of sugars in plants and is made up of a mixture of amylose and amylopectin (both polymers of glucose). ... The cells can then absorb the glucose. Starch is made up of glucose monomers that are joined by α 1-4 or α 1-6 glycosidic bonds.
Explanation:hope that helps you lots
The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
<h3>What is Enthalpy of Vaporization ?</h3>
The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
<h3>How to find the energy change from enthalpy of vaporization ?</h3>
To calculate the energy use this expression:
where,
Q = Energy change
n = number of moles
= Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) = 
= 
= 0.5 mol
Now put the values in above formula we get
[Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849
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