Answer: im thinking its gonna be d.C2H6 and also
the explanation is on the research i had did before i had answered this question so i really hope this help :)
Explanation:
Ar = van de waals forces or london forces
C
H
4
= van de waals forces or london forces
HCl=permanent dipole-dipole interactions
CO = permanent dipole-dipole interactions
HF = hydrogen bonding
N
a
N
O
3
= permanent dipole-dipole interactions
C
a
C
l
2
= van de waals forces or london forces
C. A compound !hope this helps!
Answer: The final temperature of both substances at thermal equilibrium is 301.0 K
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of gold = 31.5 g
= mass of water = 63.4 g
= final temperature = ?
= temperature of gold = 
= temperature of water = 
= specific heat of gold = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![-31.5\times 0.129\times (T_{final}-342.4)=[63.4\times 4.184\times (T_{final}-300.4)]](https://tex.z-dn.net/?f=-31.5%5Ctimes%200.129%5Ctimes%20%28T_%7Bfinal%7D-342.4%29%3D%5B63.4%5Ctimes%204.184%5Ctimes%20%28T_%7Bfinal%7D-300.4%29%5D)
The final temperature of both substances at thermal equilibrium is 301.0 K
1) Balanced chemical reaction
H2SO4 (aq) + 2NaOH(aq) ----> Na2SO4(aq) + 2 H2O(liq)
2) Molar ratios
1 mol H2SO4 : 2 mol NaOH : 1 mol Na2SO4
3) Number of moles of NaOH
M = n / V => n = M * V = 1.0 M * 0.018 liter = 0.018 mol
of NaOH.
4) Determine the number of moles of H2SO4 using proportionality
1 mol H2SO4 / 2 mol NaOH = x / 0.018 mol NaOH
Solve for x:
x = 0.018 mol NaOH * 1 mol H2SO4 / 2 mol NaOH =0.009 mol H2SO4
5) Calculate the molarity using 0.009 mol and 25 mililiters
M = n / V = 0.009 mol / 0.025 liter = 0.36 M
Answer: the concentration of the H2SO4 is 0.36 M
