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3 years ago

A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. What is the volume of the ball?

Chemistry

2 answers:

erik [133]3 years ago

4 0

Answer:

The volume of the ball is 4.82 mL (opton C)

Explanation:

Step 1: Data given

Mass of the lead ball = 55.0 grams

Density = 11.4 g/cm³

Step 2: Calculate volume of the balloon

Volume = mass / density

Volume = 55.0 g / 11.4 g/cm³

Volume = 4.82 cm³ = 4.82 mL

The volume of the ball is 4.82 mL (opton C)

Lemur [1.5K]3 years ago

3 0

Answer:

Explanation:

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3 years ago

How many Joules are released to cool 250.0 grams of liquid water from 100°C to 0°C? The specific heat of water is 4.180 J/g.C.

nlexa [21]

$\bold{\huge{\orange{\underline{ Solution}}}}$

$\bold{\underline{ Given :- }}$

We have 250g of liquid water and it needs to be cool at temperature from 100° C to 0° C
Specific heat of water is 4.180J/g°C

$\bold{\underline{ To \: Find :- }}$

We have to find the total number of joules released.

$\bold{\underline{ Let's \:Begin:- }}$

We know that,

Amount of heat energy = mass * specific heat * change in temperature

That is, 

$\sf{\red{ Q = mcΔT }}$

Subsitute the required values in the above formula :-

$\sf{ Q = 250 × 4.180 ×(0 - 100 )}$

$\sf{ Q = 250 × 4.180 × - 100 }$

$\sf{ Q = 250 × - 418}$

$\sf{\pink{ Q = - 104,500 J }}$

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

$\bold{\underline{ Now :- }}$

We have to tell whether the above process is endothermic or exothermic :-

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Endothermic process :- It is the process in which heat is absorbed .

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2 years ago

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3 years ago

If 0.500 mol neon at 1.00 atm and 273 K expands against a constant external pressure of 0.100 atm until the gas pressure reaches

trapecia [35]

Answer:

The work done on neon = -323 J

The internal energy change= -392.84 J

The heat absorbed by neon = -69.84 J

Explanation:

Step 1: Data given

Number of moles = 0.500 moles

Pressure = 1 atm

Temperature = 273 Kelvin

The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.

a) calculate the work done on neon

W = -P(V2-V1)

⇒ with P = the pressure = 0.1 atm

⇒ with V1 = the initial volume = nRTi /Pi

⇒ with V2 = the final volume = nRTf /Pf

W = -PnR((T2/P2) -(T1/P1))

⇒ with T2 = the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

⇒ with P2 = the final pressure = 0.200 atm

⇒ with P1 = the initial pressure = 1.00 atm

W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))

W = -nR*(105 - 27.3)

W= -(0.500)*(8.314)*(77.7)

W = -323 J

b) calculate the internal energy change

E = (3/2)*nRT

ΔE = Ef - Ei

ΔE =(3/2)*nR(T2-T1)

⇒ with n= number of moles = 0.500 moles

⇒ with T2 =the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

ΔE = (3/2)*(0.5)*(8.314)(210-273)

ΔE = -392.84 J

c) Calculate the heat absorbed by neon

ΔE = q + W

q = ΔE -W

⇒ with ΔE = -392.84 J

⇒ with W = -323 J

q = -392.84 J -( -323 J)

q =-392.84 J + 323 J

q = -69.84 J

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