The relation between the volume of the gas and the temperature is established by Charles's law. With a decrease in the temperature, the volume decreases by 45.7 mL. Thus, option c is correct.

<h3>What is Charle's law?</h3>

Charle's law states the direct relation present between the temperature and the volume of the gas. The law is given as:

V₁ ÷ T₁ = V₂ ÷ T₂

Given,

V₁ = 50 mL

T₁ = 303.15 K

T₂ = 277.15 K

Substituting the value the final volume is calculated as:

50 ÷ 303.15 = V₂ ÷ 277.15

V₂ = (50 × 277.15) ÷ 303.15

= 45.71 mL

Therefore, option c. 45.7 mL is the final volume.

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5 0

2 years ago

When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so

zmey [24]

Answer: 54.4 kJ/mol

Explanation:

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

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole$

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, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 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 = $50ml+50ml=100ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g$

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 = 100 g

$T_{final}$ = final temperature of water = $27.5^0C$

$T_{initial}$ = initial temperature of metal = $21.0^0C$

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

$q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C$

$q=2719.6J=2.72kJ$

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of $HCl$ :

0.05 moles of $HCl$ releases heat = 2.72 KJ

1 mole of $HCl$ releases heat = $\frac{2.72}{0.05}\times 1=54.4KJ$

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

6 0

3 years ago

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A 25.00-mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125-mL Erlenmeyer flask and 2 drops of

Semenov [28]

Answer:

0.0611M of HNO3

Explanation:

<em>The concentration of the NaOH solution must be 0.1198M</em>

<em />

The reaction of NaOH with HNO3 is:

NaOH + HNO3 → NaNO3 + H2O

<em>1 mole of NaOH reacts per mole of HNO3.</em>

That means the moles of NaOH used in the titration are equal to moles of HNO3.

<em>Moles HNO3:</em>

12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.

In 25.00mL = 0.025L -The volume of the aliquot-:

0.00153 moles HNO3 / 0.025L =

7 0

3 years ago

Rank the elements according to highest ionization energy:<br><br> Be, C, O, Ne, B, Li, F, N

aleksklad [387]

Given:
Be - Beryllium   -   9,3227
C   - Carbon - 11,2603
O   - Oxygen - 13,6181
Ne - Neon - 21,5645
B   - Boron -   8,298
Li - Lithium -   5,3917
F   - Fluorine - 17,4228
N   - Nitrogen - 14,5341

Arranged from highest ionization energy to lowest ionization energy.

Ne ; F ; N ; O ; C ; Be ; B ; Li

5 0

3 years ago

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