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Scrat [10]
4 years ago
14

6.517×10^17+4.14×10^15

Chemistry
1 answer:
Iteru [2.4K]4 years ago
5 0
Use photo math it’s rlly good
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Bonjeur monseur, la french are here! it seems you have a problem with the elements on earth. lead seems correct since it is indeed a harmful material
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A hydrogen only spectrum
nikklg [1K]

Answer:

yes

Explanation:

6 0
3 years ago
4. Consider the following data: Metal Mass (9 Cu Specilic Heat_Wg % Temperature, 0,900 0.285 these two metals are placed in cont
lana66690 [7]

The heat will flow from copper to aluminum because Cu is at higher temperature. The heat liberated is -7.60kJ

When two metals at different temperatures are kept in contact, heat flows from hotter metal to colder metal until thermal equilibrium is reached.

Here Copper is at a temperature of 60 degree Celsius and aluminum is at 40 degree Celsius. Thus, heat will flow from Cu to Al.

In order to calculate the amount of heat liberated following calculations are required.

m1=262 g

T1=87 oC

Cp=0.385 J/g oC

T2=11.8 oC

The heat liberated can be expressed as follows:

Q=mCp(T2-T1)

Q=262 g*0.385 J/goC(11.8-87)oC

Q=-7585 J

=-7.60kJ

To learn more about heat check the link below:

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4 0
1 year ago
PLZ HELP For the reaction: 2NO2(g) → N2O4(l),
Furkat [3]

Answer: \Delta H_{rxn}=-20kJ/mol-(+66kJ/mol)

Explanation:

Heat of reaction or enthalpy change is the energy released or absorbed during the course of the reaction.

It is calculated by subtracting the enthalpy of reactants from the enthalpy of products.

\Delta H=H_{products}-H_{reactants}

\Delta H = enthalpy change = ?

H_{products} = enthalpy of products

H_{reactants} = enthalpy of reactants

For the given reaction :

2NO_2(g)\rightarrow N_2O_4(l)

\Delta H=H_{N_2O_4}-2\times H_{NO_2}

\Delta H=-20kJ/mol-(+66kJ/mol)

6 0
3 years ago
A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
4 years ago
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