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Olin [163]
3 years ago
7

From top to bottom in a group of elements, ionization energies tend to...

Chemistry
1 answer:
BigorU [14]3 years ago
3 0
The answer would be B. :)
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In beta decay, what is emitted? A. an electron B. a helium-4 atom C. a neutron D. a photon E. a radionucleus
myrzilka [38]

Answer: A. an electron

<u>Beta particles are electrons or positrons (electrons with positive electric charge or antielectrons).</u> Beta decay is a type of radioactive decay in which a beta ray is emitted from an atomic nucleus.

<u>Beta decay occurs when, in an unstable nucleus with too many protons or too many neutrons, one of the protons or neutrons transforms into the other.</u> In beta minus decay, a neutron is broken down into a proton, an electron, and an antineutrino (the neutrino antiparticle, meaning it has an opposite charge to the neutrino). In beta decay plus, a prototype breaks down into a neutron, a positron and a neutrino.

4 0
2 years ago
Read 2 more answers
What is it called when scientists write out a compound such as two hydrogen and one oxygen? PLEASE HELP
schepotkina [342]
A compound is when two or more elements are joined together. The compound that is created with two hydrogens and one oxygen is h2o or water.
7 0
3 years ago
Can you please help me
rosijanka [135]

Shure what you need help with

7 0
3 years ago
How many milliliters of 4.00 M NaOH are required to exactly neutralize 50.0 milliliters of a 2.00 M solution of HNO3 ?
kramer

Answer: The volume of NaOH required is 25.0 ml

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity HNO_3 = 1

M_1 = molarity of HNO_3 solution = 2.00 M

V_1 = volume of  HNO_3 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 4.00 M

V_1 = volume of  NaOH solution =  ?

Putting in the values we get:

1\times 2.00\times 50.0=1\times 4.00\times V_2

V_2=25.0ml

Therefore, volume of NaOH required is 25.0 ml

3 0
3 years ago
For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?
harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

6 0
3 years ago
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