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jasenka [17]
3 years ago
10

Density is the _____ of a certain ______

Chemistry
1 answer:
olga2289 [7]3 years ago
3 0
I think the first blank is "solidity" and second is "object."
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Answer:

hydrgen = i think it is 4

oxygen = i think it is 3

Explanation:

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19) Which one of the following statements describes a chemical property of hydrogen gas?
Rom4ik [11]
B) hydrogen gas burns in air.

explanation: the rest are physical properties.
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When lightning strikes, nitrogen molecules, n2, and oxygen molecules, o2, in the air react to form nitrates, no3, which come dow
faltersainse [42]
The reaction for what was describe in the problem is:

N₂ + 3 O₂ --> 2 NO₃

The reactants involved are nitrogen and oxygen gas. From the word itself, oxygen is an oxidizing agent. <em>Therefore, this reaction is an oxidation reaction due to the presence of the oxidizing agent.</em>
4 0
3 years ago
When a substance is entering a phase change, the gain or loss of heat is a result of:______.
nydimaria [60]

When a substance is entering a phase change, the gain or loss of heat is a result of energy gained or lost in forming or breaking intermolecular interaction.

The constant temperatures occur when a substance is undergoing a phase transition.  If heat is removed from a substance , such as in freezing and condensation , then the process is exothermic . In this instance , heat is decreasing the speed of the molecules causing then move slower.

Example : liquid to solid and gas to liquid .

These changes release heat to the surrounding.

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4 0
2 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
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