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Alex73 [517]
3 years ago
12

Select the correct answer. Which notation is used to represent a beta particle?​

Chemistry
1 answer:
trapecia [35]3 years ago
3 0

Explanation:

I think the notation used to represent beta is B

You might be interested in
what concentration in parts per million in a solution that contains .008 grams of o2 dissolved in 1000 grams of water
julia-pushkina [17]
Ppm = mass of solute mg / mass of solvent kg

0.008 * 1000 =  8.0 mg ( solute )

1000 / 1000 = 1.0 kg (solvent )

ppm = 8 / 1

= 8.0 ppm

hope this helps!
7 0
3 years ago
Hydrogen bonds are approximately _____% of the bond strength of covalent c-c or c-h bonds.
Lelu [443]
Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.
4 0
3 years ago
Calculate the energies of the n=2 and n=3 states of the hydrogen atom in Joules per<br> atom
kondor19780726 [428]

Answer:

See Explanation

Explanation:

Positional Energy for electron as function of principle energy level (n)

=> Eₙ = -A/n²; A = 2.18x10⁻¹⁸J

Positional Energy for electron in n=2 => E₂ = -2.18x10⁻¹⁸/(2)² = -5.45x10⁻¹⁹J

Positional Energy for electron in n=3 => E₃ = -2.18x10⁻¹⁸/(3)² = -2.42x10⁻¹⁹J

ΔE(n=3→2) = -5.45x10⁻¹⁹J - (-2.42x10⁻¹⁹J) =   -3.03x10⁻¹⁹J

7 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is ________ kj. the specific heats of ice, wat
liubo4ka [24]

Answer : The enthalpy change for converting 1 mole of ice at -25.0^oC to water at 90^oC is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

(1):ice(-25^oC)\rightarrow ice(0^oC)\\\\(2):ice(0^oC)\rightarrow water(0^oC)\\\\(3):water(0^oC)\rightarrow water(90^oC)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)]

where,

\Delta H = enthalpy change

m = mass of water = 1mole\times 18g/mole=18g

c_{ice} = specific heat of ice = 2.09 J/gk

c_{water} = specific heat of water = 4.18 J/gk

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion : 0^oC=273k

T_1 = initial temperature of ice = 0^oC=273k

T_2 = final temperature of ice = -25^oC=273+(-25)=248k

T_3 = initial temperature of water = 0^oC=273k

T_4 = final temperature of water = 90^oC=273+90=363k

Now put all the given values in the above expression, we get

\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k]

\Delta H=7712.106J=7.712KJ     (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at -25.0^oC to water at 90^oC is, 7.712 KJ

3 0
3 years ago
The mass number of an element is equal to
klasskru [66]

Answer:

The mass number is defined as the total number of protons and neutrons in an atom.

Explanation:

8 0
4 years ago
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