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3 years ago

Which of these is a physical property?

Chemistry

2 answers:

Ber [7]3 years ago

6 0

Which of these is a physical property?A. Malleable

lukranit [14]3 years ago

3 0

Malleable is a physical property of matter

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Gizmo Student exploration feel the heat if a solute has really strong bonds between its particles , would you expect it to form

Troyanec [42]

Answer:

The temperature of the solute/solvent without any external effect would decrease.

Explanation:

As the bonding between the solute particles is really strong, therefore a large amount of energy is required to overcome these forces. So that the new bonding between the solute and solvent is created.

In order to achieve this, there will be a lot of energy required and that is through the heating process. So the solution will require energy so the solute will dissolve fully either by provision of external force i.e stirring or by heating.

4 0

2 years ago

At 50.14k a substance has a vapor pressure of 258.9 torr. calculate its heat of vaporization in kj/mol if it has a vapor pressur

gayaneshka [121]

T₁ = 50,14 K.
p₁ = 258,9 torr.
T₂ = 161,2 K.
p₂ = 277,5 torr.
R = 8,314 J/K·mol.
Using Clausius-Clapeyron equation:
ln(p₁/p₂) = - ΔHvap/R · (1/T₁ - 1/T₂).
ln(258,9 torr/277,5 torr) = -ΔHvap/8,314 J/K·mol · (1/50,14 K - 1/161,2 K).
-0,069 = -ΔHvap/8,314 J/K·mol · (0,0199 1/K - 0,0062 1/K).
0,0137·ΔHvap = 0,573 J/mol.
ΔHvap = 41,82 J.

5 0

2 years ago

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

7 0

2 years ago

The equilibrium constant, K, for the following reaction is 5.10X10 at 548 K. NH_CH(s) 2 NH3(E) + HC1(2) Calculate the equilibriu

Serga [27]

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = $K_c=5.10\times 10^{-6}$

Moles of ammonium chloride = 0.573 mol

Concentration of ammonium chloride = $\frac{0.573 mol}{1.00 L}=0.573 M$

$NH_4HCl(s)\rightleftharpoons 2 NH_3(g) + HCl(g)$

Initial: 0.573 0 0

At eq'm: (0.573-x) x x

We are given:

$[NH_4Cl]_{eq}=(0.573-x)$

$[HCl]_{eq}=x$

$[NH_3]_{eq}=x$

Calculating for 'x'. we get:

The expression of $K_{c}$ for above reaction follows:

$K_c=\frac{[HCl][NH_3]}{[NH_4Cl]}$

Putting values in above equation, we get:

$5.10\times 10^{-6}=\frac{x\times x}{(0.573-x)}$

$2.9223\times 10^{-6}-5.10x\times 10^{-6}=x^2$

$x^2-2.9223\times 10^{-6}+(5.10\times 10^{-6})x=0$

On solving this quadratic equation we get:

x = 0.01707 M

The equilibrium concentration of HCl is 0.01707 M.

3 0

2 years ago

What volume in liters of 5.75×10−2 M KH2AsO4 contains 6.30 g KH2AsO4?

AnnZ [28]

Answer : The volume of solution is, 0.609 L

Explanation : Given,

Mass of $KH_2AsO_4$ = 6.30 g

Molar mass of $KH_2AsO_4$ = 180 g/mole

Concentration = $5.75\times 10^{-2}M$

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }KH_2AsO_4}{\text{Molar mass of }KH_2AsO_4\times \text{Volume of solution (in L)}}$

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

$5.75\times 10^{-2}M=\frac{6.30g}{180g/mole\times \text{Volume of solution (in L)}}$

$\text{Volume of solution (in L)}=0.609L$

Therefore, the volume of solution is, 0.609 L

3 0

2 years ago