B. Breathing problems
C. Global warming
D. Soil erosion
Answer:
The percent yield of this reaction is 92.7 %
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 34.0 grams
Mass of ammonia (NH3 produced = 41.0 grams
Molar mass of N2 = 28.0 g/mol
Molar mass of NH3 = 17.02 g/mol
Actual yield of ammonia = 38 grams
Step 2: The balanced equation
N2(g) + 3H2(g) → 2NH3(g)
Step 3: Calculate moles
Moles = mass / molar mass
Moles N2 = 34.0 grams / 28.0 g/mol
Moles N2 = 1.214 moles
Step 4: Calculate moles NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 1.214 moles N2 we'll have 2* 1.214 = 2.428 moles NH3
Step 5: Calculate mass NH3
Mass NH3 = moles * molar mass
Mass NH3 = 2.428 moles * 17.02 g/mol
Mass NH3 = 41 grams
Step 6: Calculate percent yield for the reaction
Percent yield = (actuald yield / theoretical yield) * 100 %
Percent yield = (38 grams / 41 grams ) * 100 %
Percent yield = 92.7 %
The percent yield of this reaction is 92.7 %
Answer:
Explanation:
Hello,
In this case, by considering the given chemical reaction, with given mass of nitrogen, one could compute the theoretical yield of ammonia as shown below and considering their 1 to 2 molar relationship in the chemical reaction:
In such a way, the percent yield is obtained as shown below:
Best regards.
Hey there!:
E = energy gained (input) - energy lost (output)
∆E = 150J - 115 J
∆E = 35 J
Hope that helps!
The change in internal energy (?E) of the system is 35 J based on the first law of thermodynamics.
This problem can be solved using the first law of thermodynamics, which states that the change in internal energy (?E) of a system is equal to the heat added to it (Q) minus the work done by it (W) - this is expressed as ?E = Q - W.
In this particular case, the system loses 115 J of heat, so Q equals -115 J (as it's lost, it's negative), and 150J work is performed on the system which equals +150 J (as work is done on it, it is positive). Therefore, substituting these values into the formula, we get: ?E = -115 J - (-150 J) = 35 J:
So, the change in internal energy of the system is 35 J.
#SPJ3
An addition of acid increases the solubility of calcium phosphate because the hydrogen ions can easily dissociate and form salts. And example is from the reaction 3H2SO4 + Ca3(PO4)2 → 2H3PO4 + 3CaSO4. CaSO4 is the salt of sulfuric acid and calcium phosphate.