Reactivity is an impetus For which a chemical substance undergoes A chemical either by itself or with another materialWith an overall release of energy
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
Since the density of water is 1 g /mL, hence there is 100
g of H2O. So total mass is:
m = 100 g + 5 g = 105 g
=> The heat of reaction can be calculated using the
formula:
δhrxn = m C ΔT
where m is mass, C is heap capacity and ΔT is change in
temperature = negative since there is a decrease
δhrxn = 105 g * 4.18 J/g°C * (-2.30°C)
δhrxn = -1,009.47 J
=> However this is still in units of J, so calculate
the number of moles of NaCl.
moles NaCl = 5 g / (58.44 g / mol)
moles NaCl = 0.0856 mol
=> So the heat of reaction per mole is:
δhrxn = -1,009.47 J / 0.0856 mol
δhrxn = -11,798.69 J/mol = -11.8 kJ/mol
Answer:
The method is accurate in the calculation of the 
Explanation:
As a first step we have to calculate the <u>average concentration </u>of
find it by the method.

Then we have to find the<u> standard deviation:</u>

For the confidence interval we have to use the formula:
μ=Average±
Where:
t=t student constant with 95 % of confidence and 5 data=2.78
μ=
± 
upper limit: 0.84
lower limit: 0.75
If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.
The question is incomplete. Complete question is attached below:
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Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.