<span> It is important to keep the NaOH solution covered at all time because sodium hydroxide is a very good remover of Carbon dioxide from the air means sodium hydroxide absorbs the carbon dioxide from the air react with that so the concentration of your solution will also change if you uncover the NaOH.
The following reaction occurs when sodium hydroxide reacts with carbon dioxide;
</span><span>2 NaOH(aq) + CO2(g) --> Na3CO3(aq) + H2O(l) </span>
Answer is: her reasoning is flawed, because <span>Kc is very small, so the concentration of nitric(II) oxide is also very small. </span>
Balanced chemical reaction: N₂(g) + O₂(g) ⇄<span> 2NO(g).
</span>The equilibrium
constant<span> (Kc) is
a ratio of the concentration of the products (in this reaction nitrogen(II) oxide) to the concentration of the reactants (in this reaction nitrogen and oxygen):
</span>Kc = [NO]² / [N₂] · [O₂].
Kc = 4.7·10⁻³¹.
If we take equilibrium concentration of oxygen and nitrogen to be 1 M:
[N₂] = [O₂] = 1 M.
[NO] = √[N₂] · [O₂] · Kc.
[NO] = 6.855·10⁻¹⁶ M; equilibrium concentration of nitric oxide.
Answer:
3CaBr2 + 2LI3PO4 - > Ca3(PO4) 2 + 6LiBr
Explanation:
The first one I did was PO4. There are two on the right side, so I added 2 to Li3PO4 on the other side. That balanced the PO4s and then gave me 6 Lithiums so I balanced that one next on the right side. I added 6 to LiBr which balanced the Li but then gave me 6 Br, so I finished it off by adding 3 in front of CaBr2 which balanced the calcium and bromines.
Here was the process:
CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr
Balances PO4 (2on both sides)
CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr
Balances Lithiums (6 on each side)
3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr
Balances Calciums and Bromines (3 Calciums and 6 Bromines each side)
Hope this helped!
The pressure would increase. When the temperature change form cold to hot, the gas will find ways to escape from containment. Thus, if it cannot escape that pressure will keep on increasing as the temperature rises.
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O
