You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
Answer:
A scientific question is basically a question that can lead to a hypothesis to help us figure out the observation in science. I hope this helps you
Answer:- 
Explanations:- Lattice energy depends on two factors, charge and size.
High charge and small size gives higher lattice energy where as low charge and bigger size gives lower lattice energy.
in LiCl, NaCl and KCl, the anion is same and also the charges for Li, Na and K are also same. The deciding factor here is the size of cations. Since the size increases as we move down a group, the order of size of these three atoms is Li<Na<K.
The order of lattice energy is exactly opposite as it's increases as the size decreases.
Now, if we look at magnesium chloride and strontium chloride then again the anion is common but the metals have higher charge as compared to the alkali metals(Li, Na and K). So, lattice energy values must be higher for these two compounds. If we compare Mg and Sr then size of Mg is smaller and so the lattice energy would be greater for this.
Hence, the increasing order of lattice energy is
.
Answer:
T = 215.33 °C
Explanation:
The activation energy is given by the Arrhenius equation:

<u>Where:</u>
k: is the rate constant
A: is the frequency factor
Ea: is the activation energy
R: is the gas constant = 8.314 J/(K*mol)
T: is the temperature
We have for the uncatalyzed reaction:
Ea₁ = 70 kJ/mol
And for the catalyzed reaction:
Ea₂ = 42 kJ/mol
T₂ = 20 °C = 293 K
The frequency factor A is constant and the initial concentrations are the same.
Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

(1)
By solving equation (1) for T₁ we have:
Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.
I hope it helps you!