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3 years ago

If you have a gold brick that is 2cm by 3cm by 4cm and has density of 19.3g/cm3, what is its mass?

1 answer:

4 0

Volume = a x a x a

V = 2 cm x 3 cm x 4 cm => 24 cm³

Density = 19.3 g/cm³

Mass = ?

Therefore:

m = D x V

m = 19.3 x 24

m = 463.2 g

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What temperature in kelvin does 60.5 liters of sulfur dioxide occupy if there are 2.5 mol at 0.75 atm??

7nadin3 [17]

The temperature in kelvin does 60.5 liters of sulfur dioxide occupy if there are 2.5 mole at 0.75 atm is 221.07 kelvin

Explanation
This is calculated using ideal gas equation, that is PV=nRT
where, P(pressure) = 0.75 atm
V(volume) = 60.5 L
n(moles) = 2.5 mole
R( gas constant) = 0.0821 L.atm/mol.k
T(temperature =?

by making T the subject of the formula
T is therefore =Pv/nR

T= (0.75 atm x 60.5 L) / ( 2.5 molex 0.0821 L.atm/mol.K) = 221.07 kelvins

6 0

3 years ago

This sphere contains all of the frozen water on Earth

ozzi

I believe the correct answer is the geoshpere. I hope this was helpful.

8 0

4 years ago

Read 2 more answers

When two or more simple machines are combined they form a(n) ____.

Dmitry_Shevchenko [17]

Answer:

Explanation:

A compound machine is a combination of two or more simple machines.

6 0

3 years ago

Five gases combined in a gas cylinder have the following partial pressures: 3. 00 atm (N2), 1. 80 atm (O2), 0. 29 atm (Ar), 0. 1

Helen [10]

The total pressure of the gaseous mixture has been 5.37 atm. Thus, option D is correct.

The partial pressure has been defined as the pressure exerted by each gas in the mixture.

According to the Dalton's law of partial pressure, the total pressure of gas has been the sum of the partial pressure of the gases in the mixture.

The given partial pressure of gases in the mixture has been:

Partial pressure of Nitrogen, $P_N_2=3\;\rm atm$
Partial pressure of Oxygen, $P_O_2=1.80\;\rm atm$
Partial pressure of Argon, $P_A_r=0.29\;\rm atm$
Partial pressure of Helium, $P_H_e=0.18\;\rm atm$
Partial pressure of Hydrogen, $P_H=0.10\;\rm atm$

The total pressure of the gaseous mixture has been:

$P=P_N_2\;+\;P_O_2\;+\;P_A_R\;+\;P_H_e\;+\;P_H\\P=3\;+\;1.80\;+\;0.29\;+\;0.18\;+\;0.10\;\text {atm}\\P=5.37\;\rm atm$

The total pressure of the gaseous mixture has been 5.37 atm. Thus, option D is correct.

For more information about partial pressure, refer to the link:

brainly.com/question/14623719

7 0

2 years ago

The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 k, its melting temperature. the molar volume of th

kotykmax [81]

Answer:

$\Delta _{fus}H=3255.3J/mol$

$\Delta _{fus}S=7.62\frac{J}{mol*K}$

Explanation:

Hello,

Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

$\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}$

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

$p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )$

Solving for the enthalpy of fusion we obtain:

$\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol$

Finally the entropy of fusion is given by:

$\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}$

Best regards.

5 0

3 years ago