Hey there!
C₇H₉+ HNO₃ → C₇H₆(NO₂)₃+ H₂O
Our elements: C, H, N, and O.
Balance H.
Ten on the left, eight on the right. Add a coefficient of 2 in front of H₂O.
C₇H₉+ HNO₃ → C₇H₆(NO₂)₃+ 2H₂O
Balance N.
One on the left, three on the right. Add a coefficient of 3 in front of HNO₃.
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 2H₂O
This unbalanced H, let's increase the coefficient in front of H₂O from 2 to 3 to rebalance.
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 3H₂O
Balance O.
Nine on the left, nine on the right. Already balanced.
Balance C.
Seven on the left, seven on the right. Already balanced.
Our final balanced equation:
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 3H₂O
Hope this helps!
Answer:
The answer to your question is 2.3 g
Explanation:
Data
Molecule = NaOH
grams of Na in 4 g of NaOH
Process
1.- Calculate the molecular mass of NaOH
molecular mass = (1 x 23) + (1 x 16) + (1 x 1)
= 23 + 16 + 1
= 40 g
2.- Calculate the atomic mass of sodium
atomic mass = 23 g
3.- Use proportions to solve this problem
40 g of NaOH ---------------- 4 g of NaOH
23 g of Na ---------------- x
x = (4 x 23) / 40
x = 2.3 g
Answer:
The atomic mass of nitrogen is
<h2>
14.0067 u</h2>
<h2>
<em>You learn more from failure than from success</em></h2>
<h2>
Simple diffusion is defined as the movement of: A molecules from areas of higher concentration to areas of lower concentration.</h2>
Explanation:
Diffusion
It is the movement of substances from the region of higher concentration to lower concentration .
This diffusion is shown almost negligible by solids , a little by liquids and maximum by gases .
The factors on which diffusion depends :
- Inter-molecular force between particles : It is inversely proportional to diffusion .That is : more is the force , lesser the diffusion rate
- Inter-molecular space between particles : More is the space , more is the diffusion or vice versa .
- Density of substance :More is the density , less is the diffusion or vice versa .
Applications of diffusion
It helps in the movement of substance in and out from the cell and its components .
It helps in spreading the fragrance when sprayed with some perfume etc .
It helps in dissolving substance in any medium
Etc
This problem uses the relationship between Kb and the the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:
<span>
Kb = KaKb
</span><span>1.00 x 10^-14 = 7.2 x 10^-4(x)
</span><span>x = 1.39 x 10^-11
</span><span>
We now need to calculate the [OH¯] using the Kb expression:
</span>1.39 x 10^-11 = x^2 / (0.30 - x)
<span>
The denominator can be neglected. </span><span>Thus, x is 3.73 x 10^-6.
</span><span>
pOH = -log 3.73 x 10^-6 = 5.43
p</span><span>H = 14-5.43 = 8.57</span>