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3 years ago

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:

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

Hope this helps you

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Why are metals malleable?

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Answer:

b is the anwer

Explanation:

the option is the explanation

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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