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3 years ago

What are 3 mediums that mechanical waves travel through.

Chemistry

1 answer:

motikmotik3 years ago

6 0

Answer: water surf

Explanation:

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Solid phosphorus and chlorine gas react to form phosphorus pentachloride gas .write a balanced chemical equation for this reacti

EleoNora [17]

Balanced chemical equation :

2 P + 5 Cl₂ = 2 PCl₅

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7 0

3 years ago

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Answer fast and I’ll give BRAINLIEST please and thank you!! IT WOULD HELP ALOT IF U DONT RUSH AND ANSWER CORRECTLY

andrezito [222]

Answer:

i am thinking that they will be extinct

Explanation:

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4 years ago

What determines the strength of a dipole dipole force?

Reil [10]

Answer:

The more polar the molecules, the stronger the force. The more hydrogen in the molecule, the stronger the force.

Explanation:

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3 years ago

7. A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is its density

BaLLatris [955]

Answer:

$d=1.6\ g/cm^3$

Explanation:

Given that,

The dimension of a square chunk of plastic is 5 cm × 5 cm × 5 cm

Mass, m = 200 g

We need to find its density

Density = mass/volume

It will form a cube. It volume = side³

So,

$d=\dfrac{200\ g}{5^3\ cm^3}\\\\d=1.6\ g/cm^3$

Hence density is $1.6\ g/cm^3$

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4 years ago

A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t

leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = $P_1=1.0 atm$

Initial temperature of the oxygen gas = $T_1=20^oC =293.15 K$

Final pressure of the oxygen gas = $P_2=?$

Final temperature of the oxygen gas = $T_2=100^oC =373.15 K$

Using Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm$

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

$P_1V_1=nRT_1$ (ideal gas equation)

$n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol$

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

6 0

3 years ago