Answer:
115g/mol
Explanation:
To get the molar mass, we know that the it is equal to the mass divided by the number of moles. We have the mass but we do not have the number of moles.
We get this by working through the solution information. Firstly, we need to know the number of moles in 750ml for a molarity of 0.29m
Now, since 0.29 moles is present in 1000ml, x moles will be present in 750ml
The value of x is obtained as follows:
x = (750 * 0.29)/1000 = 0.2175 moles
Now since we have the number of moles, we can then obtain the molar mass.
Molar mass = mass/number of moles = 25.0g/0.2175 = 114.94 approximately 105g/mol
Answer is ₂₀⁴²Ca.<span>
In AZX notation
A<span> - Represents the </span>mass number of atom.
Mass number = number of
protons + number of neutrons.
<span> It is shown at
the </span>above of the left side<span> of the
atom.</span>
<span> Here it is </span>42.
<span> </span>Z - Represents the atomic number.<span>
Atomic number = number of protons.
</span><span>It is shown at the</span> bottom line of the left side of the atom.<span>
</span>Here it is 20.<span>
X - </span>Represents the symbol of the atom.<span>
<span> </span></span>Here</span> X is Ca.<span>
</span>
Bonds formed between atoms can be classified as ionic and covalent
Ionic bonds are formed between atoms that have a high difference in the electronegativity values.
In contrast, bonds formed between atoms that have a difference in electronegativity lower than the ionic counterparts are polar covalent bonds. If the atoms have very similar electronegativities, they form non-polar covalent bonds.
In H2S, the S atom is bonded to 2 H atoms. The electronegativity of H = 2.2 and S= 2.56. Since the difference is not high the bond formed will be covalent (polar covalent).
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:

We know that,
The relation between the
for an ideal gas are :

As we are given :



Now we have to calculate the entropy change of the gas.


(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.