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IrinaK [193]
3 years ago
8

Two cars approach each other from opposite directions each

Physics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

Answer:524 Hz

Explanation:

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A car's horn produces a constant frequency of 350 Hz as it passes by Suzy. What is the best estimate of the
Arada [10]

Answer:

330 Hz

Explanation:

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> Which number marks a crest?
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number 2 because the curve demstrates the crest GOOD LUCK i hope i got you the correct answer if not sorry

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3 years ago
4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box?
bazaltina [42]

Answer:

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V =  0.0619 m^3

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3 years ago
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Min is asked to balance the equation below.
Minchanka [31]
It is incorrect, because the identity of the original product has changed. Ca3(OH)2 does not exist! It is no longer calcium hydroxide. To balance an equation, you must manipulate the coefficients, a.k.a. the big numbers that go before reactants or products. Subscripts, the little numbers inside the reactants or products, cannot be changed without completely changing the substance.
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3 years ago
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What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

T_1= Freezing point of solution

Using the formula

3.63=0-T_1

Freezing point of water=0 degree Celsius

T_1=0-3.63=-3.63 C

Hence, the freezing point of solution=-3.63 degree Celsius

7 0
3 years ago
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