Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
Answer:
The answer is
<h2>16.8 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question
mass of object = 52.9 g
density = 3.14 g/mL
The volume of the object is
We have the final answer as
<h3>16.8 mL</h3>
Hope this helps you
It is a soft,malleable,and ductile metal
Answer:
![[H^+]=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5x10%5E%7B-13%7DM)
![[OH^-]=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.02M)
Explanation:
Hello there!
In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:
Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:
![\frac{1}{3} =\frac{x}{[Mg(OH)_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7Bx%7D%7B%5BMg%28OH%29_2%5D%7D)
Thus, x for this problem is:
![x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5BMg%28OH%29_2%5D%7D%7B3%7D%3D%5Cfrac%7B0.03M%7D%7B3%7D%5C%5C%5C%5Cx%3D%20%200.01M)
Now, according to an ICE table, we have that:
![[OH^-]=2x=2*0.01M=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2x%3D2%2A0.01M%3D0.02M)
Therefore, we can calculate the H^+, pH and pOH now:
![[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B0.02%7D%3D5x10%5E%7B-13%7DM)
Best regards!
