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3 years ago

15

Balloon A started with a volume of 1.76 L. The temperature of the room the balloon was in was 295 K. The balloon was heated to a

temperature of 253.15 K. What is the new volume?

1 answer:

valentina_108 [34]3 years ago

4 0

Answer:

V₂ = 1.5 L

Explanation:

Given data:

Initial volume of balloon = 1.76 L

Initial temperature = 295 K

Final temperature = 253.15 K

Final volume = ?

Solution:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1.76 L ×253.15 K / 295 K

V₂ = 445.54 L.K /295 K

V₂ = 1.5 L

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Based on the activity series, which one of the reactions below will occur (Spontaneous)? A. Mn (s) + NiCl2 (aq) → MnCl2 (aq) + N

kap26 [50]

Answer:

Mn (s) + NiCl2 (aq) → MnCl2 (aq) + Ni

Explanation:

The order of displacement of metals from aqueous solution by another metal is defined by the activity series of metals.

The activity series arranges metals in order of reactivity and increasing electrode potentials. The less negative the electrode potential of a metal is, the less reactive it is and the lower it is found in the activity series.

Nickel has a less negative electrode potential than manganese hence it is displaced from an aqueous solution of its salt by manganese spontaneously.

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3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago

Can someone help me please and please show work

FinnZ [79.3K]

Answer:

Average of the trial is: 288.50 C

Percent Error: 3.83%

Explanation:

(291 + 287 + 295 + 281) : 4 = 288.50 C

Average: 288.50 C

Percent Error: {(300 - 288.50) : 300} x 100% = 3.83 %

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3 years ago

Is Amplitude a measure of a waves? A. speed B. energy C. distance D. frequency

iren2701 [21]

the answer is C. Distance

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3 years ago

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In trying to control fall armyworms in crops, an Agriculture extension officer applied cypermethrin which was prepared by dissol

sveta [45]

Answer:

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.0086

Explanation:

Mole fraction remains a sort of concentration. It indicates:

moles of solute / (moles of solute + moles of solvent)

Moles of solute / Total moles.

Solute: Cypermethrin → C₂₂H₁₉Cl₂NO₃

Solvent: Water (PM = 18g/mol)

We calculate moles from solvent: 1000g /18 g/mol = 55.5 moles

We calculate PM for C₂₂H₁₉Cl₂NO₃

12g/mol . 22 + 1g/mol . 19 + 35.45 g/mol . 2+ 14g/mol + 16g/mol . 3 = 416 g/m

Moles of solute: 200 g / 416g/mol = 0.481 moles

Total moles: 0.481 + 55.5 = 55.98 moles

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.481 moles / 55.98 moles = 0.0086

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3 years ago