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skad [1K]
3 years ago
9

3. Which of the following is NOT a

Chemistry
1 answer:
erik [133]3 years ago
4 0
Moons is there only jaun
You might be interested in
Map
grin007 [14]

Answer: 1650 hope i got it in time.

Explanation:

3 0
3 years ago
• An element in group 2, period 4 has similar properties with
babymother [125]

Answer:  Elements in Group 2

Explanation:  The periodic table was arranged by Dmitri Mendeleev specifically  around similarites in their chemical behaviors.  He found that as atomic number increases, at some point an element starts to react in a manner similar to a previous one.  When that happened, he would place the larger element under the smaller one, and eventually noticed a periodicity in the table.  Elements in a column (Groups) had similiar chemical properties.  We know today that these similarities are due to the electron configuration, and that these configurations repeat themselves.  He left gaps in the table when he could find an existing element with properties similar to others in that group.  I big leap of faith, but it worked.  Elements for those missing boxes were eventually discovered.

7 0
3 years ago
5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wav
torisob [31]

Answer: The range of wavelengths of light that can be used to cause given phenomenon is 8.953 \times 10^{21} m.

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

E = \frac{hc}{\lambda}

where,

E = energy

h = Planck's constant = 6.625 \times 10^{-25} Js

c = speed of light = 3 \times 10^{8} m/s

Substitute the values into above formula as follows.

E = \frac{hc}{\lambda}\\222000 J = \frac{6.625  \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m

Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is 8.953 \times 10^{21} m.

7 0
3 years ago
If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees
telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁  +  q₂  =  0

n  =  moles of NH₄NO₃  =  8.00 g NH₄NO₃  ×  1 mol NH₄NO₃/80.0 g NH₄NO₃          

∴ n =   0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m  =  mass of solution  =  1000.0 g + 8.00 g  =  1008.0 g

q₂  =  mcΔT  = 58.0 g  ×  4.184 J°C⁻¹  g⁻¹  × ((20.39-21)°C) = -2570.19 J

q₁  +  q₂  =  0.100 mol  ×ΔHsoln  – 2570.19 J  =  0

ΔHsoln  =  +2570.19 J  /0.100 mol  =  +25702 J/mol  =  +25.7 kJ/mol

7 0
4 years ago
We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a
sukhopar [10]

Answer:

See Explanation

Explanation:

\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) \frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0
3 years ago
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