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3 years ago

MAGNESIUM (Mg) +FLUORINE (F) =

Chemistry

1 answer:

bazaltina [42]3 years ago

4 0

Answer:

MgF2 magnesium fluoride

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Which describes the products created from fission reactions? Question 3 options:

Ipatiy [6.2K]

<u>Answer:</u> The correct answer is option A.

<u>Explanation:</u>

Nuclear fission reactions are a type of nuclear reactions in which larger nuclei breaks apart into two or more smaller fragment releasing alpha, gamma of beta particles.

There are 3 types of particles that can be released during this process:

1. Alpha particles: These particles are released when a nuclei undergoes alpha-decay process.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

2. Beta particles: These particles are released when a nuclei undergoes beta-minus decay process.

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

3. Gamma radiations: these radiations are released when an unstable nuclei gives off excess energy by a process of spontaneous electromagnetic process.

$_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma$

Hence, any of these particles can be released during the process of fission reaction with smaller atoms.

Therefore, the correct answer is option A.

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3 years ago

An electron is moving northward in a magnetic field. the magnetic force on the electron is toward the northeast. what is the dir

sweet-ann [11.9K]

The direction of the magnetic field would be going up.

8 0

2 years ago

2H2O + 2K (s) → 2KOH (1) + 2H2 (g) is an example of which type of reaction?

AlexFokin [52]

Its a single Displacement reaction

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2 years ago

Read 2 more answers

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

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3 years ago

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stellarik [79]

Answer:

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Explanation:

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2 years ago