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2 years ago

What type of radioactivity occurs in nuclear reactor if a neutron and uranium are the typical reactants

Chemistry

1 answer:

prisoha [69]2 years ago

6 0

Answer:

Fission. When a large fissile atomic nucleus such as uranium-235 or plutonium-239 absorbs a neutron, it may undergo nuclear fission. The heavy nucleus splits into two or more lighter nuclei, (the fission products), releasing kinetic energy, gamma radiation, and free neutrons.

Explanation:

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Give the reason for the following; 1. Iodimetric titrations are usually performed in neutral or mildly alkaline (pH 8) to weakly

Agata [3.3K]

Answer:

Explanation:

blach bal

5 0

2 years ago

43 mg = [?]g <br>A. 0.043 g <br>B. 4.3 g <br>C. 4300 g <br>D. 43,000 g

Ymorist [56]

Answer:

Option A (0.043 g) is the correct answer.

Explanation:

Given:

= 43 mg

As we know,

$1 \ mg = \frac{1}{1000} \ g$

then,

⇒ $43 \ mg = \frac{43}{1000} \ g$

$= 0.043 \ g$

Thus, the above is the correct alternative.

4 0

2 years ago

Which two types of information are written in an element's box in the periodic table?

GenaCL600 [577]

Answer:

Yes it is B,D.

Explanation:

Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups.

8 0

2 years ago

What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i

Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314 (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0

2 years ago

A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike and the rider is 85 kg. Find the kinetic ener

krek1111 [17]

KE=3070.625 J

Height = 3.686 m

<h3>Further explanation</h3>

mass of bike+rider=85 kg

velocity = 8.5 m/s

$\tt KE=\dfrac{1}{2}mv^2\\\\KE=\dfrac{1}{2}\times 85\times 8.5^2\\\\KE=3070.625~J$

Conservation of energy :

(KE+PE)₁ (downhill) = (KE+PE)₂ (up the hill)

PE₁=0⇒h=0

KE₂=0⇒v=0(stop), so equation becomes :

KE₁=PE₂

$\tt KE_1=3070.625\\\\KE_1=mgh_2\\\\3070.625=85\times 9.8\times h\Rightarrow h=3.686~m$

5 0

2 years ago