Repeated trials and replication are the same thing.

이

Nataliya [291]

The daughter isotope : Radon-222 (Rn-222).

<h3>Further explanation</h3>

Given

Radium (Ra-226) undergoes an alpha decay

Required

The daughter nuclide

Solution

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,

alpha α particles ₂He⁴
beta β ₋₁e⁰ particles
gamma particles ₀γ⁰
positron particles ₁e⁰
neutron ₀n¹

The decay reaction uses the principle: the sum of the atomic number and mass number before and after decay are the same

Radium (Ra-226) : ₈₈²²⁶Ra

Alpha particles : ₂⁴He

So Radon-226 emits alpha α particles ₂He⁴ , so the atomic number decreases by 2, mass number decreases by 4

The reaction :

₈₈²²⁶Ra ⇒ ₂⁴He + ₈₆²²²Rn

4 0

2 years ago

Which step is part of a scientific investigation?

Elis [28]

Steps of a scientific investigation include identifying a research question or problem, forming a hypothesis, gathering evidence, analyzing evidence, deciding whether the evidence supports the hypothesis, drawing conclusions, and communicating the results.

3 0

2 years ago

Why does pressure vary in the atmosphere

Digiron [165]

Pressure varies in the atmosphere because air molecules are being pulled down towards the center of the earth-hope this helps!

7 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.