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NemiM [27]
3 years ago
7

One mole of water is equivalent to 18 grams of water. A glass of water has a mass of 200 g. How many moles of water is this?

Physics
2 answers:
lawyer [7]3 years ago
6 0

Answer:

11.11 moles

Explanation:

1 mole of water is equivalent to 18 g of water.

18 gram of water contains = 1 mole of water

1 gram of water contains = 1 / 18 mole of water

So, 200 g of water contains = 200 / 18 moles of water

                                               = 11.11 moles of water

Thus, 200 g of water contains 11.11 moles.

Black_prince [1.1K]3 years ago
4 0
200g*1 mole/ 18g=11.1 moles There are 11.1 moles of water. 
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How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

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3 years ago
What would happen to the fossil when weathering occurs. Would it survive ? ASAP
Kamila [148]
Yes it would servive beacuse the fossils are dug down into the rockand unable to move
6 0
3 years ago
Consider a series RLC circuit where R = 855 Ω and C = 6.25 μF. However, the inductance L of the inductor is unknown. To find its
sashaice [31]

Answer:

L= 0.059 mH

Explanation:

Given that

R = 855 Ω and C = 6.25 μF

V= 84 V

Frequency

ω = 51900 1/s

We know that

\omega=\sqrt{\dfrac{1}{LC}}

L=Inductance

C=Capacitance

ω =angular Frequency

ω² L C =1

(51900)² x L x 6.25 x 10⁻⁶ = 1

L= 5.99 x 10⁻⁵ H

L= 0.059 mH

6 0
3 years ago
An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

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