Fter a radioactive atom decays, it is the same element that it was before with no measurable change in mass. Which kind of decay has occurred, and how do you know?

Due to high amount of dissolved oxygen in cold water, goldfish have a better chance of survival.

what is dissolved oxygen?

Dissolved oxygen is the amount of oxygen that is present in the water bodies.

As dissolved oxygen is very much important for the aquatic animals because their survival depends on the water. When we increases the temperature of water then the amount of dissolved oxygen decreases which cuts off the supply of oxygen to the aquatic animals. That's why every aquatic specie has a better chance of survival in the cold water, due to presence of the abundant dissolved oxygen.

Hence, goldfish has a better survival in cold water because of dissolved oxygen.

To learn more about dissolved oxygen, visit below link:

brainly.com/question/26073928

The correct statement regarding the first ionization energy and electronegativity values in group 15 is as follows:

Further Explanation:

The amount of energy needed for the removal of the most loosely bound electron from a neutral isolated gaseous atom is termed as ionization energy, represented by IE. It depends on the ease of electron removal from the neutral atoms. If the electrons are removed easily, ionization energy will be less and vice-versa.

Ionization energy is called the first ionization energy when the first electron is removed from the atom. It is shown by . Similarly, if the second electron is removed, ionization energy becomes the second ionization energy .

The tendency of any element for electron attraction towards itself in a chemical bond is known as electronegativity. More the attracting tendency of the atom for electron, higher will be its electronegativity and vice-versa.

Atomic number and number of shells increase while going down group 15. Due to this, atomic size increases in the group from top to bottom. This increase in size results in weaker attractions between the outermost electrons and the atomic nucleus. So electrons are removed easily and therefore the first ionization energy decreases down this group.

Since atomic size increases from top to bottom of group 15, the attraction between the atomic nucleus and the electrons decreases. Therefore electronegativity also decreases down this group.

Therefore both the first ionization energy and electronegativity decrease down group 15.

Learn more:

1. Rank the elements according to first ionization energy: brainly.com/question/1550767
2. Arrange the following elements from greatest to least tendency to accept an electron: brainly.com/question/2107660

Answer details:

Grade: Senior School

Chapter: Periodic classification of elements

Subject: Chemistry

Keywords: ionization energy, first ionization energy, electronegativity, attraction, group 15, decrease.

The answer is has a Moh's hardness of 1.

The enthalpy for the reaction, ΔH rxn is 255.95 kJ/mol

From the question,

We are to calculate the change in enthalpy for the reaction

CH₄(g) + NH₃(g) → HCN(g) +3H₂(g)

From the given reactions

Reaction 1: N₂(g) + 3H₂(g) → 2NH₃(g) Change in enthalpy: -91.8 kJ/mol

Reaction 2: C(s, graphite) + 2H₂(g) → CH₄(g) Change in enthalply: -74.9 kJ/mol

Reaction 3: H₂(g) +2C(s, graphite) +N₂(g) → 2HCN (g) Change in enthalpy: +270.3 kJ/mol

First, flip reactions 1 and 2 to get reaction 4 and 5 respectively

Reaction 4: 2NH₃(g)  → N₂(g) + 3H₂(g)                 ΔHo : 91.8 kJ/mol

Reaction 5: CH₄(g) → C(s, graphite) + 2H₂(g)       ΔHo : 74.9 kJ/mol

Now, multiply reactions 4 and 3 by half (1/2) to get 6 and 7 respectively

Reaction 6: NH₃(g)  → ¹/₂N₂(g) + ³/₂H₂(g)                           ΔHo : 45.9 kJ/mol

Reaction 7: ¹/₂H₂(g) +C(s, graphite) +¹/₂N₂(g) → HCN (g)   ΔHo : +135.15 kJ/mol

Now,

Add reactions 5, 6, and 7 together  

Reaction 5: CH₄(g) → C(s, graphite) + 2H₂(g)                    ΔHo : 74.9 kJ/mol

Reaction 6: NH₃(g)  → ¹/₂N₂(g) + ³/₂H₂(g)                           ΔHo : 45.9 kJ/mol

Reaction 7: ¹/₂H₂(g) +C(s, graphite) +¹/₂N₂(g) → HCN(g)    ΔHo : +135.15 kJ/mol

-------------------------------------------------------------------------------------------------------------

CH₄(g) + NH₃(g) → HCN(g) + 3H₂(g)                             ΔH rxn = 255.95 kJ/mol

Hence, the enthalpy for the reaction, ΔH rxn is 255.95 kJ/mol

Learn more here: brainly.com/question/13779366

Answer:

255.8 kj/mol

Explanation:

So this is a Hess' Law problem, the CH₄ (g) + NH₃ (g) --> HCN (g) + 3H₂ (g) is what we want the other reactions to reflect. I usually set up problems like these like this in order to determine which reaction needs a coefficient change:

N₂ + 3H₂ --> 2NH₃             (ΔH=-91.8)

C + 2H₂ --> CH₄                 (ΔH=-74.9

H₂ + 2C + N₂ --> 2HCN     (ΔH=270.3)

CH₄ + NH₃ --> HCN + 3H₂

(I left out the states because it'll make the math easier) So, we want things to cancel out, meaning some of the reactants and products need to change places in order to do so. For the first reaction, we'd want to multiply the coefficients by in order to have it cancel out with the other reactions. For the third reaction, we'd want to we'd want to switch the products/reactants and multiply the coefficients by . Keep in mind whatever we do to the equation, we do to the ΔH. Should look like:

N₂ + H₂ --> NH₃             (ΔH=-45.9)

C + 2H₂ --> CH₄                (ΔH=-74.9)

HCN --> H₂ + C + N₂     (ΔH=-135)

CH₄ + NH₃ --> HCN + 3H₂

Everything cancels, so that means we can add all the ΔH, which should be -255.8 kj/mol, but we also change the sign in order to reflect what's happening in the reaction. (Sorry this is so long)