Question:
how many significant figures are in the measurement 0.0056 ke?
a. four
b. two
c. five
d. three
answer:
b. two
Answer:
True! Under most conditions, they are usually the same.
This is because to balance out the negativity or positivity of an atom, the opposite joins in.
Hope this helps.
Answer:
0.258 mg of iron remains.
Explanation:
To solve this problem we can use the formula
M₂ = M₀ * 
Where M₂ is the mass remaining, M₀ is the initial mass, and t is time in days.
Using the data given by the problem:
M₂ = 2.000 mg * 
M₂ = 0.258 mg
Answer:
The reaction is an exothermic reaction .
The value of q is -880 kilo Joules.
Explanation:
Exothermic reactions are defined as the reactions in which energy of reactants is more than the energy of the products. In these reactions, energy is released by the system. The total enthalpy of the reaction 
comes out to be negative.
According to reaction, 2 moles of ammonia when reacts with 3 moles of nitrous oxide to give 4 moles of nitrogen gas and 3 moles of water vapor, along with release of 880 kJ of heat energy.
Since, heat is evolved during the course of reaction which means that reaction is an example of Exothermic reaction.
The reaction is an exothermic reaction .
The value of q is -880 kilo Joules.
<h3>Answer:</h3>
The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.
<h3>Explanation:</h3>
Lets take start with the melting point of both compounds.
n-Butane = - 140 °C
Trimethylamine = - 117 °C
Intermolecular Forces in n-Butane:
As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.
Intermolecular Forces in Trimethylamine:
Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.
