It’s b and could I have the Brianliest plzzzz
Answer:
B.) More intensely
Explanation:
If you immerse a lightstick in hot water, the chemical reaction will speed up. The stick will glow much more brightly but will wear out faster too.
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Answer:
Al 72.61%
Mg 27.39%
Explanation:
To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.
If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.
To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.
Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g
The mass of magnesium is 0.0381 * 24 = 0.9144g
We can now calculate the mass percentage.
The total mass is 0.9144 + 2.4246 = 3.339g
% mass of Al = 2.4246/3.339 * 100 = 72.61%
% mass of Mg = 0.9144/3.39 * 100 = 27.39%
Answer:
3.24 × 10^5 J/mol
Explanation:
The activation energy of this reaction can be calculated using the equation:
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
Where; Ea = the activation energy (J/mol)
R = the ideal gas constant = 8.3145 J/Kmol
T1 and T2 = absolute temperatures (K)
k1 and k2 = the reaction rate constants at respective temperature
First, we need to convert the temperatures in °C to K
T(K) = T(°C) + 273.15
T1 = 325°C + 273.15
T1 = 598.15K
T2 = 407°C + 273.15
T2 = 680.15K
Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)
ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4
7.831 = Ea(2.417 × 10^-5)
Ea = 3.24 × 10^5 J/mol
