What is the coefficient for H_2O when the following combustion reaction of a fat
ID: 473740 • Letter: W
Question
What is the coefficient for H_2O when the following combustion reaction of a fatty acid is C_18 H_36O_2 + O_2 rightarrow CO_2 + H_2O 1 18 9 26 27 All of the following are properties of Tungsten, which one is a chemical property? Tungsten can be hammered into a thin sheet. At 60 degree C a sheet of tungsten crumbles to a gray powder. Tungsten boils at 5555 degree C. When a bar of tungsten is bent, it emits an audible "cry". Tungsten erodes when added to hydrochloric acid, and a clear gas forms. A mixture of O_2 gas and an unknown gas are placed in a container with a pinhole in its side If the oxygen gas is found to leak at a rate 2.14 times faster than the unknown gas, which of these could be the unknown gas? Cl_2 SF_6 Kr UF_6 X_e How many grams of NaOH are required to make 350mL of 0.400 M NaOH? 45.7 5.60 35.0 44.2 1.75 Based on the activity series, which one of the reactions below will occur? Pb(s) + NiI_2(aq) rightarrow Pbl_2(aq) + Ni(s) SnBr_2(aq) + Cu(s) rightarrow CuBr2(aq) + Sn(s) Fe(s) + ZnCl_2(aq) rightarrow FeCl_2(aq) + Zn(s) Mn(s) + NiCl_2(aq) rightarrow MnCl_2(aq) + Ni(s) None of the reactions will occur.Explanation / Answer
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What is the coefficient for H2O when the following combustion reaction of fatty acid is balanced?
Answer is B
C18H36O2 + 26O2 18CO2 + 18H2O
32]
All of the following are properties of Tungsten which one is a chemical property?
Answer is E
Tungsten erodes when added to HCl and a clear gas forms.
33]
A mixture of O2 gas and an unknown gas are placed in a container with a pinhole in its side. If the oxygen gas is found to leak at a rate 2.14 times faster than the unknown gas which of these could be the unknown gas
Answer is B
SF6
The rate of leakage through a pinhole is, according to Graham's Law, inversely proportional to the square root of the molecular weight of the gas.
Oxygen gas has a MW of 2 x 16 = 32
If R = rate of leakage, M = Molecular Weight we have:
R2 = R1(M2/M1)^(1/2)
M2 = M1 *(R1/R2)^2
M1 for oxygen gas is 2 x 16 =32, M2 = 32*2.14^2 =146.5
UF6 has M > 200
Molecular weight of: Xe = 131
SF6 = 32 + 6*19 =146
Cl2 = 2*35.5 = 71
Kr = 84
The only gas that fits is gas SF6
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Answer is B
Moles of solute = Molarity × Liters of solution
Multiply 0.450 M by 0.200:
0.400 mol/ 1 L×0.350 L=0.14 mol
To obtain the mass of solute, we will need to the molar mass of NaOH, which is 40.00 g/mol:
Finally, multiply the number of moles by 40.00 g/mol
= 0.14mol×40.00g/1mol
= 5.60g
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