Answer:
569K
Explanation:
Q = 3.5kJ = 3500J
mass = 28.2g
∅1 = 20°C = 20 + 273 = 293K
∅2 = x
c = 0.449
Q = mc∆∅
3500 = 28.2×0.449×∆∅
3500 = 12.6618×∆∅
∆∅ = 3500/12.6618
∆∅ = 276.4220
∅2 - ∅1 = 276.4220
∅2 = 276.4220 + ∅1
∅2 = 276.4220 + 293
∅2 = 569.4220K
∅2 = 569K
bromine
Explanation:
halogens are a group of elemnts simlar to eachother
flourine, chlorine, and bromine
Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the 
orbit of hydrogen atom.
Formula used :
where,
= energy of 
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:
Energy of n = 2 in an hydrogen atom:
Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.
The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV
Balanced equation for the above reaction is as follows;
Mg(OH)₂ + 2HCl ---> MgCl₂ + 2H₂O
stoichiometry of Mg(OH)₂ to MgCl₂ is 1:1
mass of Mg(OH)₂ reacted - 1.82 g
number of moles of Mg(OH)₂ - 1.82 g/ 58.3 g/mol = 0.0312 mol
number of Mg(OH)₂ moles reacted - number of MgCl₂ moles formed
number of MgCl₂ moles formed - 0.0312 mol
mass of MgCl₂ formed - 0.0312 mol x 95.2 g/mol = 2.97 g
mass of MgCl₂ formed - 2.97 g
Your control group would be the batteries since you CONTROL what brand you're using, for which one lasts the longest...aren't you suppose to figure that out when you do the experiment?
