1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO
solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
or 1 / 0.65 = x / 16
24.6 g is the answer.
if 1 gram of oxygen requires 1.65 grams of Mg
then 16 grams of oxygen will require 16 ( 1.65) or 26.4 grams.
Answer:
HCl is not a catalyst because these are not used up during the chemical reactions.
Explanation:
Hello there!
In this case, according to the performed experiments, it is possible for us to realize that HCl cannot be a catalyst for this reaction because it is used up during the reaction. This is explained by the fact that catalyst are able to return to the original form once the reaction has gone to completion; this is the example of palladium in the hydrogenation or dehydrogenation of hydrocarbons depending on the case. Moreover, we know that the catalysts increase the reaction rate because they decrease the activation energy of the reaction and therefore the student observed such increase.
Best regards!
Answer:
11552.45 years
Explanation:
Given that:
Half life = 5730 years
Where, k is rate constant
So,
The rate constant, k = 0.00012 years⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 0.00012 years⁻¹
Initial concentration = 160.0 counts/min
Final concentration = 40.0 counts/min
Time = ?
Applying in the above equation, we get that:-
Answer:
P₂ = 0.09 atm
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 0.225 L
Initial pressure = 338 mmHg (338/760 =0.445 atm)
Initial temperature = 72 °C (72 +273 = 345 K)
Final temperature = -15°C (-15+273 = 258 K)
Final volume = 1.50 L
Final pressure = ?
Solution:
P₁V₁/T₁ = P₂V₂/T₂
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L
P₂ = 25.83 atm .L. K / 293 K . L
P₂ = 0.09 atm
<h3>1</h3>
Species shown in bold are precipitates.
- Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
- Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃
- Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
- Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
- Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃
- Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
- Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃
- Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
- Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃
<h3>2</h3>
A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.
Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.
<h3>3</h3>
Compare the first and last row:
Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.
As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.
<h3>4</h3>
Compare the second and third row:
Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.
