Answer:
1. NaN₃(s) → Na(s) + 1.5 N₂(g)
2. 79.3g
Explanation:
<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>
NaN₃(s) → Na(s) + 1.5 N₂(g)
<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>
First, we have to calculate the moles of N₂ from the ideal gas equation.
The moles of NaN₃ are:
The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:
Answer:
Water gradually degrades them until the metal parts rust and the other materials deteriorate.
Explanation:
Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>
1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.
(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>
Answer:
It means the chemical entity is a radical
Explanation:
When we talk of unsaturation, we are referring to the number of pi-bonds in a chemical entity. The alkane, alkene and alkyne organic family are used to as common examples to explain the term unsaturation.
While alkynes have 3 bonds, it must be understood that they have 2 pi bonds only and as such their degree of saturation is two.
In the case of an alkene, there is only one single pi bond and as such the degree of unsaturation is 1.
Now in this case, we have a fractional 0.5 degree of unsaturation alongside the 3 to make a total of 3.5. So what’s the issue here?
The fractional part shows that the chemical entity we are dealing with here is a radical. While the integer 3 shows that there are 3 pi-bonds, the half pi bond remaining tells us that there is a missing electron on one of the atoms involved in the chemical bonding and as such, the 1/2 extra degree of unsaturation tends to tell us this.
Kindly recall that a radical is a chemical entity within which we have at the least an unpaired electron.
