The temperature at which the process be spontaneous is calculated as follows
delta G = delta H -T delta S
let delta G be =0
therefore delta H- T delta s =0
therefore T= delta H/ delta S
convert 31 Kj to J = 31 x1000= 31000 j/mol
T=31000j/mol /93 j/mol.k =333.33K
Answer:
3. The temperatures of the two substances equalize.
Explanation:
- As two objects at different temperatures are placed in contact, heat is transferred from the warmer to the cooler object until the temperature of the two objects be the same.
- The amount of heat that is transferred from the warmer object is equal to the amount of heat that is transferred into the cooler object.
- This is in agreement with the law of conservation of energy.
- <em>So, the right choice is: 3. The temperatures of the two substances equalize. </em>
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Answer : The molecular weight of a gas is, 128.9 g/mole
Explanation : Given,
Density of a gas = 5.75 g/L
First we have to calculate the moles of gas.
At STP,
As, 22.4 liter volume of gas present in 1 mole of gas
So, 1 liter volume of gas present in 
mole of gas
Now we have to calculate the molecular weight of a gas.
Formula used :
Now put all the given values in this formula, we get the molecular weight of a gas.
Therefore, the molecular weight of a gas is, 128.9 g/mole
The value 6.0 x 10^3- 2.3 × 10^3 in scientific notation is 3.7 × 10^3.
<h3> What is scientific notation?</h3>
Scientific notation is a way to write very large or very small numbers so that they are easier to read and work with.
You express a number as the product of a number greater than or equal to 1 but less than 10 and an integral power of 10 .
<h3>Why it is used? </h3>
There are two reasons to use scientific notation.
- The first is to reveal honest uncertainty in experimental measurements.
- The second is to express very large or very small numbers so they are easier to read.
Given,
= 6.0 x 10^3- 2.3 × 10^3
= (6.0 - 2.3) × 10^3
= 3.7 × 10^3
Thus, we find that the value 6.0 x 10^3- 2.3 × 10^3 in scientific notation is 3.7 × 10^3.
learn more about scientific notation :
brainly.com/question/18073768
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Answer:
Some of the physical changes used by the industrial chemist in order to identify it is by scratching it with other metals in order to find the hardness of it. Trying to deform it in order to find the malleability, and to heat it and measure the temperature in order to find the melting point.
Some of the chemical changes used by the industrial chemist in order to identify it is by inserting it in water to observe that whether it reacts with it or not, if the reaction is violent, then the metal belongs to either group I or group II. The other method is to insert it in acids of distinct strength and to observe its reaction. The metals belonging to the second group react briskly with acids. The other metals react gradually with acids and others are almost inert.
