Answer:
the normality of the given solution is 0.0755 N
Explanation:
The computation of the normality of the given solution is shown below:
Here we have to realize the two sodiums ions per carbonate ion i.e.
N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)
= 0.1886eq ÷ 0.2500L
= 0.0755 N
Hence, the normality of the given solution is 0.0755 N
Answer:
1
Explanation:
For non metals to attain a noble gas configuration, they gain the number of electrons needed to attain the noble gas configuration of the noble gas at the end of their periods. This means that these non metals would only take up the configuration of the last element on their periods which of course is always a noble gas.
The last element on the hydrogen period or more conservatively the only other element on the hydrogen period is helium, with an atomic number of 2. The atomic number is the number of protons in he nucleus of an atom. For an electrically neutral atom, the number of electrons equal the number of protons.
Hence we can deduce that helium has 2 electrons while hydrogen has one electron. Thus for it to attain the configuration of helium, it just needs to gain one more electron
Answer:
7,94 minutes
Explanation:
If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>
For the zero-order reactions, concentration-time equation can be written as follows:
[A] = - Kt + [Ao]
where:
- [A]: concentration of the reactant A at the <em>t </em>time,
- [A]o: initial concentration of the reactant A,
- K: rate constant,
- t: elapsed time of the reaction
<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>
Data:
K = 4.2 ×10−3atm/s,
[A]o=[HBr]o= 2 atm,
[A]=[HBr]=0 atm (all HBr(g) is gone)
<em>We clear the incognita :</em>
[A] = - Kt + [Ao]............. Kt = [Ao] - [A]
t = ([Ao] - [A])/K
<em>We replace the numerical values:</em>
t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes
So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).
Answer : The amount of carbon dioxide produced is, 197.12 grams.
Explanation : Given,
Moles of ethanol = 2.24 mole
Molar mass of carbon dioxide = 44 g/mole
The balanced chemical reaction will be,
First we have to calculate the moles carbon dioxide.
From the balanced chemical reaction, we conclude that
As, 1 mole of ethanol react to give 2 moles of carbon dioxide
So, 2.24 mole of ethanol react to give 
moles of carbon dioxide
Now we have to calculate the mass of carbon dioxide.
Therefore, the amount of carbon dioxide produced is 197.12 grams.
<h2>Natural Abundance for 10B is 19.60%</h2>
Explanation:
- The natural isotopic abundance of 10B is 19.60%.
- The natural isotopic abundance of 11B is 80.40%.
- The isotopic masses of boron are 10.0129 u and 11.009 u respectively.
For calculation of abundance of both the isotopes -
Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.
Determining it as an equation -
10x + 11y= 10.8
x+y=1 (ratio)
10x + 10y = 10
By taking the denominator away from the numerator
we get;
y = 0.8
x + y = 1
∴ x = 0.2
To get percentages we need to multiply it by 100
So, the calculated abundance is 80% for 11 B and 20% 10 B.
