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3 years ago

6

Suppose that the root-mean-square velocity vrms of water molecules (molecular mass is equal to 18.0 g/mol) in a flame is found t

o be 1150 m/s. What temperature does this represent

1 answer:

Zolol [24]3 years ago

4 0

Answer:

$T=954.41\ K$

Explanation:

The expression for the root mean square speed is:

$C_{rms}=\sqrt {\dfrac {3RT}{M}}$

R is Gas constant having value = 8.314 J / K mol

M is the molar mass of gas

Molar mass of water vapor = 18.0 g/mol = 0.018 kg/mol

Temperature = ?

$C_{rms}=1150$ m/s

$1150=\sqrt{\frac{3\times 8.314\times T}{0.018}}$

$1322500=\frac{24.942T}{0.018}$

$T=954.41\ K$

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if a plot weight (in g) vs. volume (in ml) for a metal gave the equation y= 13.41x and r^2=0.9981 what is the density of the met

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ANS: density = 13.41 g/ml

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Consider the reaction: HCN (aq) + H2O (l) ⇄ CN- (aq) + H3O + (aq) Which of the following statements will decrease the amount of

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2 years ago

At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall

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Answer:

concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

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concentration of $[CO_2]$ = 0.4442 M

Explanation:

Equation for the reaction:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Concentration of $CO_2_{(g)$ = $\frac{2.3}{4.9}$ = 0.469

For our ICE Table; we have:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Initial 0.469 0 0

Change - 2x +2x +x

Equilibrium (0.469-2x) 2x x

K = $\frac{[CO]^2[O]}{[CO_2]^2}$

K = $\frac{[2x]^2[x]}{[0.469-2x]^2}$

$4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}$

Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²

$4.1*10^{-6} = \frac{2x^3}{(0.938)}$

$2x^3 =3.8458*10^{-6$

$x^3 =\frac{3.8458*10^{-6}}{2}$

$x^3=1.9229*10^{-6$

$x=\sqrt[3]{1.9929*10^{-6}}$

x = 0.0124

∴ at equilibrium; concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

concentration of $[CO]$ = 2x = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of $[CO_2]$ = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

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3 years ago