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nadya68 [22]
3 years ago
11

True or False Helium belongs to Noble Metals

Chemistry
2 answers:
Lostsunrise [7]3 years ago
6 0

Answer:

This answer is "True"

Vlada [557]3 years ago
6 0
The answer would be *true*
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238 93 Np → 0 -1 e+ ?​
Sergeeva-Olga [200]

Answer:

²³⁸₉₃Np  → Pu₉₄²³⁸ + ⁰₋₁e

Explanation:

²³⁸₉₃Np  → Pu₉₄²³⁸ + ⁰₋₁e

Beta radiations:

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiation

⁴₆C → ¹⁴₇N + ⁰₋₁e

The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.

6 0
3 years ago
How do I convert Kilometers to miles? And Kg to pounds?
d1i1m1o1n [39]

there is a google calculator for this, but i don't know the exact formula.

8 0
3 years ago
Which of the following words mean ä substance formed by chemically combining two or more elements
kogti [31]

Answer:

B. Compound

Explanation:

Hope I helped

3 0
3 years ago
Read 2 more answers
why can you find different features at an oceanic-continental convergent zone than those found at a continental-continental conv
arlik [135]
Continental plates are much thicker that Oceanic plates. At the convergent boundaries the continental plates are pushed upward and gain thickness. The rocks and geological layers are much older on continental plates than in the oceanic plates. The Continental plates are much less dense than the Oceanic plates.
7 0
2 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
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