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1 year ago

Calculate the root mean square velocity, in m/s, of carbon dioxide molecules at 100.0

Chemistry

1 answer:

spayn [35]1 year ago

8 0

Answer:

459.8m/s

Explanation:

T=Temperature in Kelvin

M=molar mass in kg/mol

Molar mass of CO2,

mm=44.01g/mol

=0.04401kg/mol

T =100.0C

= (100.0 +273) K

=373K

So,

vrms= sqrt(3×R×T/M)

vrms=sqrt(3×8.314×373.0/0.04401)

=459.8m/s

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2 years ago

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2 years ago

What is true of a reaction that has reached equilibrium? The rate of the forward reaction is greater than the rate of the revers

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2 years ago

Read 2 more answers

When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from

RideAnS [48]

Answer : The enthalpy of reaction $(\Delta H_{rxn})$ is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of $AgNO_3$ and $HCl$ .

$\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole$

$\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole$

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

$AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)$

As, 1 mole of $AgNO_3$ react with 1 mole of $HCl$ to give 1 mole of $AgCl$

So, 0.005 mole of $AgNO_3$ react with 0.005 mole of $HCl$ to give 1 mole of $AgCl$

The moles of AgCl formed = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

$q=m\times C\Delta T=m\times C \times (T_2-T_1)$

where,

q = heat

C = specific heat capacity = $4.18J/g^oC$

m = mass = 100 g

$T_2$ = final temperature = $24.21^oC$

$T_1$ = initial temperature = $23.40^oC$

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

$q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC$

$q=338.58J$

Now we have to calculate the enthalpy of the reaction.

$\Delta H_{rxn}=\frac{q}{n}$

where,

$\Delta H_{rxn}$ = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

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

$\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole$

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction $(\Delta H_{rxn})$ is, 67.716 KJ/mole

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2 years ago

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(c)

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(d)

$\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\ \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}$

(e)

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