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Andreas93 [3]
3 years ago
12

A wave is moving towards shore with a velocity of 5.0 MS if it’s frequency is 2.5 Herz what is the wave length

Chemistry
1 answer:
EastWind [94]3 years ago
8 0

Answer:

<h2>2 m</h2>

Explanation:

The wavelength of a wave can be found by using the formula

\lambda =  \frac{c}{f}  \\

where

c is the speed of the wave

f is the frequency

From the question

c = 5 m/s

f = 2.5 Hz

We have

\lambda =  \frac{5}{2.5}  = 2 \\

We have the final answer as

<h3>2 m </h3>

Hope this helps you

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Answer:

b is the anwer

Explanation:

the option is the explanation

4 0
3 years ago
In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0
sp2606 [1]

Answer:

The value of K_p for this reaction at 1200 K is 4.066.

Explanation:

Partial pressure of water vapor at equilibrium = p^o_{H_2O}=15.0 Torr

Partial pressure of hydrogen gas at equilibrium = p^o_{H_2}=?

Total pressure of the system at equilibrium P = 36.3 Torr

Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:

P=p^o_{H_2O}+p^o_{H_2}

p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr

3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)

The expression of K_p is given by:

K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}

K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066

The value of K_p for this reaction at 1200 K is 4.066.

6 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
Who was the first person to discover the existence of electrons​
k0ka [10]
Electron was discovered by J. J. Thomson in 1897 when he was studying the properties of cathode ray.
4 0
3 years ago
Read 2 more answers
The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
2 years ago
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