Answer:
0.1 mol
Explanation:
Step 1: Given data
- Volume of the solution of sodium hydroxide (V): 200 cm³
- Molar concentration of sodium hydroxide (C): 0.5 mol/dm³
Step 2: Convert "V" to dm³
We will use the conversion factor 1 dm³ = 1000 cm³.
200 cm³ × 1 dm³/1000 cm³ = 0.200 dm³
Step 3: Calculate the moles (n) of NaOH
The molarity of the NaOH solution is equal to the moles of NaOH divided by the volume of solution.
C = n/V
n = C × V
n = 0.5 mol/dm³ × 0.200 dm³ = 0.1 mol
Answer:
this is soloution of example ..
Answer:
The appropriate alternative is option C "".
Explanation:
The given value is:
Energy of photon,
E =
As we know,
h =
Now,
The frequency of the light will be:
⇒
On substituting the given values, we get
⇒
On solving, we get
⇒
This hypothetical process would produce actinium-230.
<h3>Explanation</h3>
An alpha decay reduces the atomic number of a nucleus by two and its mass number by four.
There are two types of beta decay: beta minus β⁻ and beta plus β⁺.
The mass number of a nucleus <em>stays the same</em> in either process. In β⁻ decay, the atomic number <em>increases </em>by one. An electron e⁻ is produced. In β⁺ decay, the atomic number <em>decreases </em>by one. A positron e⁺ is produced. Positrons are antiparticles of electrons.
β⁻ are more common than β⁺ in decays involving uranium. Assuming that the "beta decay" here refers to β⁻ decay.
Gamma decays do not influence the atomic or mass number of a nucleus.
Uranium has an atomic number of 92. 238 is the mass number of this particular isotope. The hypothetical product would have an atomic number of 92 - 2 ⨯ 2 + 1 = 89. Actinium has atomic number 89. As a result, the product is an isotope of actinium. The mass number of this hypothetical isotope would be 238 - 2 ⨯ 4 = 230. Therefore, actinium-230 is produced.
The overall nuclear reaction would involve five different particles. On the reactant side, there is
On the product side, there are
- one actinium-230 atom,
- two alpha particles (a.k.a. helium-4 nuclei),
- one electron, and
- one gamma particle (a.k.a. photon).
Consider: what would be the products if the nucleus undergoes a β⁺ decay instead?