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3 years ago

Substances X and Y are both nonpolar. If the volatility of X is higher than that of Y, what is the best explanation?

Physics

2 answers:

Studentka2010 [4]3 years ago

8 0

The answer is,

D. Y’s molecules experience stronger London dispersion forces than X’s molecules.

lesya692 [45]3 years ago

4 0

I believe the correct answer from the choices listed above is the last option. If the volatility of X is higher than that of Y, then Y’s molecules experience stronger London dispersion forces than X’s molecules. All molecules has london dispersion forces. Also, the stronger the bond, the harder it is to volatilize. Hope this answers the question.

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a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an

11111nata11111 [884]

When looking for distance you multiply speed by time

So 15 x 2 = 30

30 is the distance between his house and school

6 0

3 years ago

The mole is 6.02 x 10 23 particles. If a person masses out the correct molar mass in grams for a substance then she would have a

Leto [7]

Answer:

1 mole of H2O is 18 grams (2 g H + 16 g Oxygen)

36 / 18 = 2

So 2 moles = 2 * 6.02E23 = 12.04E23 = 1.204E24

7 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

Serhud [2]

For the answer to the question above, first find out the gradient.

m = rise/run 
=(y2-y1)/(x2-x1) 

the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" 
(x1,y1) = (3,53) 
(x2,y2) = (6,62) 

sub values back into the equation 
m = (62-53)/(6-3) 
m = 9/3 
m = 3 

now we use a point-slope form to find the the standard form 
y-y1 = m(x-x1) 
where x1 and y1 are any set of point given 
y-53 = 3(x-3) 
y-53 = 3x - 9 
y = 3x - 9 + 53 
y = 3x + 44 

y is the velocity of the car, x is the time.
I hope this helps.

4 0

3 years ago

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

grandymaker [24]

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, $\theta=35^{\circ}$

Time of flight :

$t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s$

Horizontal distance traveled is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m$

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

6 0

2 years ago