Which best describes a difference between prokaryotic and eukaryotic cells

What is the velocity of a wave traveling north with a frequency of 75 hertz and a wavelength of

Vanyuwa [196]

Answer:

v = 450 m/s

Explanation:

Given data:

Frequency = 75 Hz

Wavelength = 6 m

Velocity = ?

Solution:

Velocity is the product of frequency and wavelength.

v = f × λ

v = 75 Hz × 6 m

Hz = s⁻¹

v = 75 s⁻¹ × 6 m

v = 450 m/s

3 0

2 years ago

Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

The reaction quotient (Q) before the reaction is 0.32

4 0

2 years ago

it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to

schepotkina [342]

Answer:

The right solution is " $8.5\times 10^{-19} \ joule$ ".

Explanation:

As we know,

1 mole electron = $6.023\times 10^{23} \ no. \ of \ electrons$

Total energy = $513 \ KJ$

= $513\times 1000 \ joule$

For single electron,

The amount of energy will be:

= $\frac{513\times 1000}{(6.023\times 10^{23})}$

= $8.5\times 10^{-19} \ joule$

8 0

2 years ago

Which of the following is not a possible termination step in the free radical chlorination of methane? Group of answer choices ∙

vagabundo [1.1K]

Answer: first option is not a termination

∙CH3 + Cl2 → CH3Cl + Cl∙

Explanation:

Since a radical is formed as part of the product it means it's a propagation step and not a termination step, at termination no free radical exist as product

7 0

3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

7 0

2 years ago