Identify find a functional groups as indicated

Write a nuclear equation to describe the neutron induced fission of Pu-239 to form Kr-89 and Ce-149 (this fission is induced by

zlopas [31]

Answer:

$\rm 2_{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, 3_{0}^{1}\text{n}; three$

Explanation:

If the fission is induced by two neutrons, the unbalanced equation is

$2\rm _{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, x_{0}^{1}\text{n}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation .

Balancing the superscripts, we get

2 + 239 = 89 + 149 + x

241 = 238 + x

x = 3

The balanced nuclear equation is

$\rm 2_{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, \mathbf{3}_{0}^{1}\text{n}$

$\text{The fission produces } \boxed{\textbf{three}} \text{ neutrons}$

4 0

3 years ago

Please I need this help me

BaLLatris [955]

Answer:

first of all is that really chemistry

3 0

3 years ago

How much calcium hypochlorite (65% strength) is needed to make 200 L of 2% hypochlorite solution? (Assume that a 1% solution is

g100num [7]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

65 x V1 = 2 x 200 L

V1 = 6.15 L

5 0

3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.