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larisa [96]
3 years ago
10

A metal melts at 450 degrees celsius. Is this property of the metal classified as chemical or physical?

Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0
It's a physical property because it doesn't change the element of the metal.
does this make sense?
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Không tính toán cụ thể mà lý luận để :
kotykmax [81]

Answer:

c) Tìm những chất có hàm lượng đồng bằng nhau trong 4 chất : CuO, Cu2O, CuS, Cu2S.

Explanation:

3 0
2 years ago
How many moles of O2 are needed to react completely with 35.0 mol C2h2
Wittaler [7]

Hey There!:


2 C2H2 + 5 O2 = 4 CO2 + 2 H2O


2 moles C2H2 ----------- 5 moles O2

35.0 moles C2H2 ------- moles O2


moles O2 = 35.0 * 5 / 2


moles O2 = 87.5 moles


hope this helps!

6 0
3 years ago
Which equation represents a conservation of atoms?
weqwewe [10]

Answer:

4) 4Fe + 3O₂ → 2Fe₂O₃

Explanation:

4Fe + 3O₂ → 2Fe₂O₃

In this equation the numbers of atoms are same in both side. There are four iron and six oxygen atoms are present on left and right side of equation. That's why atoms are conserved. This equation completely followed the law of conservation of mass.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

8 0
2 years ago
Explain why the melting point of calcium is high?
DanielleElmas [232]
It takes more energy to breakdown the bonds
5 0
2 years ago
Read 2 more answers
One hundred cubic meters of carbon dioxide initially at 150 oC and 50 bar is to be isothermally compressed in a frictionless pis
gulaghasi [49]

Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3

Image of page 3

Explanation:As revealed above, the stimuli connections are clearly stated

4 0
2 years ago
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