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2 years ago

11

....Which is the best classification for West Nile virus?

2 answers:

Igoryamba2 years ago

8 0

Answer:

D.

Explanation:

Emerging infectious disease

Luba_88 [7]2 years ago

5 0

Answer:

The answer is F

Explanation: Only because Henrieta wasn't the sister that got kicked out

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What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

2 years ago

he first-order rate constant for the gas-phase decomposition of dimethyl ether, (CH3)2O → CH4 + H2 + CO is 3.2 ✕ 10−4 s−1 at 450

seropon [69]

Answer:

0.290 atm is the pressure of the system after 7.7min

Explanation:

The general first-order rate constant is:

ln [A] = -kt + ln [A]₀

<em>Where [A] is concentration of A in time t,</em>

<em>K is rate constant, 3.2x10⁻⁴s⁻¹</em>

<em>[A]₀ is initial concentration = 0.336atm.</em>

<em />

7.7 min are:

7.7min * (60s / 1min) = 462s

Solving:

ln [A] = -kt + ln [A]₀

ln [A] = -<em>3.2x10⁻⁴s⁻¹*462s</em> + ln [0.336atm]

ln [A] = -1.238

[A] =

<h3>0.290 atm is the pressure of the system after 7.7min</h3>

<em />

6 0

2 years ago

During which time interval does the substance exist as both a liquid and a solid

Inessa [10]

The Chemistry Regents is one of the four science Regents exams. The other three are Earth Science, Living Environment, and Physics. You'll need to pass at least one of these four exams to graduate high school.

8 0

2 years ago

A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem

Leto [7]

Answer : The final temperature of the copper is, $87.29^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat gained = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = $0.0923cal/g^oC$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25^oC$

Now put all the given values in the above formula, we get the final temperature of copper.

$299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)$

$T_{final}=87.29^oC$

Therefore, the final temperature of the copper is, $87.29^oC$

4 0

2 years ago

Which is the correct Lewis structure for fluorine, which is a group 7A element?

Yuliya22 [10]

The correct Lewis structure for Fluorine is A.

4 0

2 years ago

Read 2 more answers