B.) Because they both are Acids
Hope this helps!
Because of differences in molecular structure, the empirical formula remains different between hydrocarbons; in linear, or "straight-run" alkanes, alkenes and alkynes, the amount of bonded hydrogen lessens in alkenes and alkynes due to the "self-bonding" or catenation of carbon preventing entire saturation of the hydrocarbon by the formation of double or triple bonds.
<span>This inherent ability of hydrocarbons to bond to themselves is referred to as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane, and in rarer cases, arenes such as benzene. This ability comes from the fact that bond character between carbon atoms is entirely non-polar, in that the distribution of electrons between the two elements is somewhat even due to the same electronegativity values of the elements (~0.30), and does not result in the formation of an electrophile.
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Here is the formula for density:
Density (D) = Mass (M) divided by Volume (V)
So you would do D = 6.147 divided by 9.3
As an as answer you would get: 0.6609677419g/cm^3
Additional information:
The formula for volume is:
V = M divided by D
The formula for Mass is:
M = D times V
I hope this helps :)
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
<span>9.40x10^19 molecules.
The balanced equation for ammonia is:
N2 + 3H2 ==> 2NH3
So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794
Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol
Moles H2 = 4.72 x 10^-4 g / 2.01588 g/mol = 2.34140921086573x10^-4 mol
Moles NH3 = 2.34140921086573x10^-4 mol * (2/3) = 1.56094x10^-4 mol
Now to convert from moles to molecules, just multiply by Avogadro's number:
1.56094x10^-4 * 6.0221409x10^23 = 9.400197448261x10^19
Rounding to 3 significant figures gives 9.40x10^19 molecules.</span>
