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uysha [10]
3 years ago
9

SOMEONE PLEASE HELP HURRY

Chemistry
1 answer:
melomori [17]3 years ago
4 0

I think the answer is A. I know it's Planet X.

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If x is a string, then x = new String("OH"); and x = "OH"; will accomplish the same thing. Group of answer choices True False
Vinvika [58]

Answer:

True is the correct answer to the above question.

Explanation:

  • If x is a string then it can be assigned by the help of two ways in java:
  • By the help of constructor:- When we write " x = new String("OH");", then it will create a pass a string "OH" into the constructor. It is because the String is a class in java and x is an object created by the constructor of the String class.
  • With the help of assigning: The "x= OH", which assigns the value of x which is an object of String class it can also use the constructor to initialize the "OH" string on the class.
  • The above question states the two scenarios which are defined above. Hence the question statement is true.
5 0
3 years ago
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What is an element?
const2013 [10]

Answer:

B) a substance made of just one kind of atom that cannot be broken down into something simpler

Explanation:

Elements are what you find on the periodic table of elements.

Take oxygen, for example. You can have one or multiple atoms, but if they are all oxygen atoms it's an element.

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3 years ago
Help on chart please!!! (Will mark brainliest)
Nata [24]

Answer:

if you don't understand ask for help thats what i do

Explanation:

7 0
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why dont trees spontaneously catch on fire and burn all the time ? why dont fires stop immediately as soon as they start ?
Kamila [148]

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3 years ago
A compression, at a constant pressure of 140 kPa, is performed on 4.0 moles of an ideal monatomic gas (Cv = 3/2 R). The compress
kobusy [5.1K]

<u>Answer:</u> The change in internal energy of the gas is 29.414 kJ.

<u>Explanation:</u>

To calculate the temperature of the gas at different volumes, we use ideal gas equation:

PV=nRT

  • When volume = 0.26m^3

We are given:

Conversion used:  1m^3=1000L

P=140kPa\\V=0.23m^3=260L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}

Putting values in above equation:

140kPa\times 260L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_i\\\\T_i=1095.06K

  • When volume = 0.12m^3

We are given:

P=140kPa\\V=0.12m^3=120L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}

Putting values in above equation:

140kPa\times 120L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_f\\\\T_f=505.41K

  • To calculate the change in internal energy, we use the equation:

\Delta U=nC_v\Delta T=nC_v(T_f-T_i)

where,

\Delta U = change in internal energy = ?

n = number of moles = 4.0 mol

C_v = heat capacity at constant volume = \frac{3}{2}R

T_f = final temperature = 1095.06 K

T_i = initial temperature = 505.41 K

Putting values in above equation, we get:

\Delta U=4\times \frac{3}{2}\times 8.314J/K.mol\times (505.41-1095.06)\\\\\Delta U=29414.1J

Converting this value in kilojoules, we use the conversion factor:

1 kJ = 1000 J

So, 29414.1J=\frac{1kJ}{1000J}\times 29414.1J=29.414kJ

Hence, the change in internal energy of the gas is 29.414 kJ.

6 0
4 years ago
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