CHEMISTRY MIDDLE SCHOOL

Substances which will glow in the dark after being exposed to sunlight are:luminescent
phosphorescent
radioactive
photosensitive

Answers

Answer 1
Answer: luminescent I'm pretty sure :)

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How many electrons does N 3-have?​

Answers

Answer: 10 electrons

Explanation:

N represents Nitrogen. Nitrogen has an atomic number of 7, this means in ground state it has 7 electrons also.

But N-3, means Nitrogen has gained 3 more electrons. So, we have 10 electrons

Please tell if this right!if not please answer with correct answers

Answers

Yes I’m. 80% sure that it’s right

Answer:

Yes, you are correct

Explanation:

Nice job :)

A solution labeled "0.105 M NaOH" would contain ______________ moles of NaOH in each liter of solution.

Answers

Answer:

A solution labeled "0.105 M NaOH" would contain 0.105 moles of NaOH in each liter of solution.

Explanation:

The concentration of a solution in Molarity (M) stands for the number of moles of that substance contained in 1 L of solution.

Molarity = Concentration in mol/L = (Number of moles of solute) ÷ (Volume of solution in L)

Molarity = Concentration in mol/L = 0.105 M = 0.105 mol/L

Number of moles of solute = ?

Volume of solution in L = 1 L

0.105 = Number of moles of solute × 1

Number of moles of solute = 0.105 mole

Hence, a solution labeled "0.105 M NaOH" would contain 0.105 moles of NaOH in each liter of solution.

Hope this Helps!!!

Consider the following intermediate chemical equations.What is the enthalpy of the overall chemical reaction

Image below

Answers

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

P_4(s)+6Cl_2\rightarrow 4PCl_3  \Delta H_1=-2439kJ (1)

4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)  \Delta H_2=3438kJ (2)

Net chemical equation:

PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H=? (3)

Adding 1 and 2 we get,

4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2\Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ   (4)

Now dividing equation (4) by 4, we get

PCl_5(g)\rightarrow PCl_3(g)+Cl_2

\Delta H=(\Delta H_3)/(4)=(1000kJ)/(4)=250kJ   (4)

Lewis dot structures explicitly show which of the following about a molecule?The total electron configuration of each atom in the molecule


The three-dimensional geometry of the atoms in the molecule


The ionic charge on each atom within the molecule


The way atoms share electrons to be connected to one another within the molecule

Answers

Answer:

The total electron configuration of each atom in the molecule

Explanation:

Electron distribution into energy levels or sublevels  of atoms can be shown with different notations or electronic configuration. One of them is the lewis dot structure.

The Electron Dot Notation or Lewis Dot structure  is a notation which shows only the chemical symbol of the element surrounded by dots to represent the valence electrons. The chemical symbol stands for the nucleus and all electrons except the valence ones. The dots are arranged at the four sides of the symbol in pairs called lone pairs Those that are unpaired are called odd electrons.

Using enthalpies of formation, calculate H.

Answers

ΔH° = -851.5 kJ/mol given that

\begin{array}{cc}\textbf{Species}&{\bf {\Delta H_f\textdegree{}}}\n \text{Fe}_2\text{O}_3\;(s) & -824.2\;\text{kJ}\cdot\text{mol}^(-1)\n\text{Al}_2\text{O}_3\;(s) & -1675.7\;\text{kJ}\cdot\text{mol}^(-1)\end{array}

(Source: Chemistry Libretexts.)

Explanation

Refer to a thermodynamic data table for the standard enthalpy of formation for each species.

Don't be alerted if the data for Al (s) and Fe (s) are missing. Why?

  • The standard enthalpy of formation of a substance measures the ΔH required to form each mole of it from the most stable allotrope of its elements under STP.
  • Both Al (s) and Fe (s) are already the most stable form of their element under STP (note that the state symbol matters.) There's no need to form them again.

As a result, \Delta H_f\textdegree{} = 0 for both Al (s) and Fe (s).

\displaystyle \Delta H_{\text{rxn}}\textdegree{} = \text{Sum of }\Delta H\text{ for all }\textbf{Product} - \text{Sum of }\Delta H\text{ for all }\textbf{Reactant}}\n\phantom{\Delta H_{\text{rxn}}\textdegree{}} = (1* \Delta H_f\textdegree{}(\text{Al}_2\text{O}_3\;(s)) + 1* \Delta H_f\textdegree{}(\text{Al}\;(s)) \n \phantom{\Delta H_{\text{rxn}}\textdegree{}=}-(1* \Delta H_f\textdegree{}(\text{Fe}_2\text{O}_3\;(s)) + 1*\Delta H_f\textdegree{}(\text{Fe}\;(s))

\Delta H_{\text{rxn}}\textdegree{}} = (1 * (-1675.7)) - (1*(-824.2)) = -851.5\;\text{kJ}\cdot\text{mol}^(-1).

The number "1" here emphasizes that in case there are more than one mole of any species in one mole of the reaction, it will be necessary to multiply the \Delta H_f\textdegree{} of that species with its coefficient in the equation.
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