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3 years ago

Which clue can be used to identify a chemical reaction as a replacement reaction?

2 answers:

4 0

"The reaction has two reactants with ions that seem to switch places" can be used to identify a chemical reaction as a replacement reaction.

myrzilka [38]3 years ago

3 0

Answer: Option (a) is the correct answer.

Explanation:

When only one element gets displaced upon chemical reaction between a compound and an element then it is known as a single replacement reaction.

For example, $Fe_{2}O_{3}(s) + 2Al(s) \rightarrow Al_{2}O_{3}(s) + 2Fe(l)$

On the other hand, when two compounds chemically react together and their ions get replaced by each other then it is known as a double replacement reaction.

For example, $AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)$

Thus, we can conclude that the reaction which has two reactants with ions that seem to switch places clue can be used to identify a chemical reaction as a replacement reaction.

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At take off a plane flies 100 km north before turning east to fly 200 km east. How far is its destination from where the plane t

andrew-mc [135]

First the plane turns 100 km North, and than 200 km East. Since both the directions are perpendicular to each other, therefore we can apply the Pythagoras theorem to calculate the distance between the destination and the point where plane took off

=100^{2}+200^{2}

D=223.60 km=224 km

Therefore, The destination is 224 km from where the plane took off

8 0

3 years ago

Write fulk form of FPS?<br>

zzz [600]

Answer:

<u>Foot per second. Foot-pound-second system. Frames per second, the frequency (rate) at which consecutive images (frames) appear on a display.</u>

Explanation:

4 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

You can increase the vapor pressure of a liquid by:

irga5000 [103]

Answer: Option (c) is the correct answer.

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

When we increase the temperature of a liquid substance then there will occur an increase kinetic energy of the molecules. As a result, they will move readily from one place to another.

Hence, liquid state of a substance will change into vapor state of the substance. This means that an increase in temperature will lead to an increase in vapor pressure of the substance.

Thus, we can conclude that you can increase the vapor pressure of a liquid by increasing temperature.

4 0

3 years ago

Motor oil , with a viscosity of 0 . 250 Ns / m2 , is flowing through a tube that has a radius of 5 . 00 mm and is 25 . 0 cm long

Thepotemich [5.8K]

Answer:

1.1775 x 10^-3 m^3 /s

Explanation:

viscosity, η = 0.250 Ns/m^2

radius, r = 5 mm = 5 x 10^-3 m

length, l = 25 cm = 0.25 m

Pressure, P = 300 kPa = 300000 Pa

According to the Poisuellie's formula

Volume flow per unit time is

$V=\frac{\pi \times Pr^{4}}{8\eta l}$

$V=\frac{3.14 \times 300000\times \left ( 5\times 10^{-3} \right )^{4}}{8\times 0.250\times 0.25}$

V = 1.1775 x 10^-3 m^3 /s

Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.

3 0

3 years ago