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tangare [24]
2 years ago
12

HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP MEEEEE

Chemistry
1 answer:
devlian [24]2 years ago
5 0

still need help with this?

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A certain isotope has 53 protons 78 neutrons and 53 electrons. What is its atomic number? What is the mass number of this atom?
statuscvo [17]

Answer:

  1. 53 protons
  2. 131g
  3. Iodine
  4. Halogens

Explanation:

atomic no. = no. of protons

= 53 proton

mass = no. of protons + no. of

neutrons

= 53 + 78

= 131

5 0
3 years ago
What is a solution considered to be if it has a ph lower than 7?
wariber [46]
Anything less that 7, would be a base.
6 0
3 years ago
A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm
VARVARA [1.3K]

<u>Answer:</u> The nuclear equations for the given process is written below.

<u>Explanation:</u>

The chemical equation for the bombardment of neutron to U-238 isotope follows:

_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

The chemical equation for the first beta decay process of _{92}^{239}\textrm{U} follows:

_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta

The chemical equation for the second beta decay process of _{93}^{239}\textrm{Np} follows:

_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta

Hence, the nuclear equations for the given process is written above.

6 0
3 years ago
Wsphorus-SL 3. In the isotope B-11, what does the 11 represent?​
gregori [183]

Answer:

Here boron-11 means the name of the element is boron and the mass number is 11

Explanation:

8 0
3 years ago
Sodium has______ Valence electrons and bismuth has ________ electrons
Triss [41]

Answer:

Sodium has 1 valence electron and bismuth has 83 electrons (5 valence electrons)

Explanation:

5 0
3 years ago
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