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3 years ago

In order for the wavelength of a wave to increase, the frequency must....

Chemistry

1 answer:

olya-2409 [2.1K]3 years ago

8 0

The answer to this is C.

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When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago

What is the answer?

Solnce55 [7]

Answer: D

Explanation: because of because

5 0

3 years ago

Wave Properties

konstantin123 [22]

Answer:

f = 80 flutter/ second

T= 0.0125 sec

Explanation:

Given data:

Fluttering of wings = 4800 time per minute

Frequency of wings = ?

Period for one wing in sec = ?

Solution:

There are 60 seconds in one minute, thus frequency of wings

4800/60 = 80 flutter per second

Period:

T = 1/f

T = 1/80

T= 0.0125 sec

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2 years ago

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5.) If 6 moles of Oxygen Gas (O2) are produced in the reaction below, how

Marat540 [252]

Answer:

im really sorry but i dont no☹

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2 years ago

Which coefficients balance this equation?

kow [346]

The answer is c. 1,2,1,2

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3 years ago

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