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Firlakuza [10]
3 years ago
9

Describe the properties of a solid.​

Chemistry
1 answer:
Hatshy [7]3 years ago
5 0

Answer:

1.Solids have definate shape

2.Solids have definate volume.

3.Solids have definate made

4.Solids do not flow

You might be interested in
Give the value of the quantum number ℓ, if one exists, for a hydrogen atom whose orbital angular momentum has a magnitude of 6√(
MakcuM [25]

Answer:

For this angular momentum, no quantum number exist

Explanation:

From the question we are told that

   The magnitude of the angular momentum is  L  = 6\sqrt{\frac{h}{2 \pi} }

The generally formula for Orbital angular momentum is mathematically represented as

           L  = \sqrt{(l * (l + 1)) } *  \frac{h}{2 \pi}

Where l is the quantum number

now  

We can look at the given angular momentum in this form as

      L  = 6\sqrt{\frac{h}{2 \pi} }    =  \sqrt{36}  * \sqrt{\frac{h}{2 \pi}} }

comparing this equation to the generally equation for Orbital angular momentum

     We see that there is no quantum number that would satisfy this equation

5 0
3 years ago
What is the concentration of hydrogen ions in a solution of KOH with a pOH of 1.72?
bezimeni [28]

Answer:

[h+]= 5.25e-13

Explanation:

pH= 14-1.72= 12.28

[h+]= 5.25e-13

3 0
3 years ago
Which reaction is a decomposition reaction? A. 2KClO3 → 2KCl + 3O2 B. 4Na + O2 → 2Na2O C. ZnS + 3O2 → 2ZnO + 2SO2 D. 2NaBr + CaF
Vedmedyk [2.9K]

Answer:

  • <em><u>Option A. </u></em><u><em>2KClO₃ → 2KCl + 3O₂</em></u>

Explanation:

There are five basic general types of chemical reactions:

  • Decomposition reactions
  • Synthesis or combination reaction
  • Single replacement reactions
  • Double replacement reactions
  • Combustion reactions

The given reactions are:

  • <u>2KClO₃ → 2KCl + 3O₂</u>

Which is, indeed, a decomposition reaction because the reactant, KClO₃, is a single substance that undergoes a reaction in which it yields two new substances, known as products: KCl and O₂.

  • <u>4Na + O₂ → 2Na₂O</u> is a synthesis or combination reaction because two reactants, Na and O₂, combine for the formation of one single new product, Na₂O.

  • <u>ZnS + 3 O₂ →  2ZnO + 2SO₂ </u>is a single replacement reaction because oxygen is replacing Zn and S in ZnS to form ZnO and plus SO₂.

  • <u>2NaBr + CaF₂ → 2NaF + CaBr₂ </u>is a double replacement reaction because two  ions (Br⁻ from NaBr and F⁻ from CaF₂) are exchanging places with other two ions (Na⁺ from NaBr and Ca²⁺ from CaF₂) two form two new ionic compounds (NaF and CaBr₂).
4 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
4 years ago
What is the total energy change for the following reaction:CO+H2O-CO2+H2
Alekssandra [29.7K]

Answer:

\large \boxed{\text{-41.2 kJ/mol}}

Explanation:

Balanced equation:    CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

(a) Enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

4 0
3 years ago
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