I think the correct answer would be NH4ClO4. It would produce a basic solution since as it dissociates into ions OH ions are being formed as the concentration of this increases, the pH of the solution would increase as well making it basic. Hope this helps.
<u>Answer:</u> The number of moles of HI in the solution is
moles.
<u>Explanation:</u>
We are given:

To calculate the concentration of a substance, we use the equation:
......(1)
- Concentration of ammonia:
![[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L](https://tex.z-dn.net/?f=%5BNH_3%5D%3D%5Cfrac%7B0.405mol%7D%7B4.90L%7D%3D0.083mol%2FL)
- Concentration of ammonium iodide:
![[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L](https://tex.z-dn.net/?f=%5BNH_4I%5D%3D%5Cfrac%7B1.45mol%7D%7B4.90L%7D%3D0.30mol%2FL)
For the given chemical reaction:

The expression of
for above equation follows:
![K_c=\frac{[HI][NH_3]}{[NH_4I]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5BNH_3%5D%7D%7B%5BNH_4I%5D%7D)
Putting values in above equation, we get:
![7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}](https://tex.z-dn.net/?f=7.0%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BHI%5D%5Ctimes%200.083%7D%7B0.30%7D)
![[HI]=2.53\times 10^{-4}](https://tex.z-dn.net/?f=%5BHI%5D%3D2.53%5Ctimes%2010%5E%7B-4%7D)
Calculating the moles of hydrogen iodide by using equation 1, we get:

Hence, the number of moles of HI in the solution is
moles.
Answer:
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation:
Step 1: Data given
The temperature of a gas = 25.0°C
AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.
The volume reduces to 7.88 L but the temperature stays constant.
Step 2: Boyle's law
(P1*V1)/T1 = (P2*V2)/T2
⇒ Since the temperature stays constant, we can simplify to:
P1*V1 = P2*V2
⇒ with P1 = the initial pressure 667 torr
⇒ with V1 = the initial volume = 10.0 L
⇒ with P2 = the final pressure = TO BE DETERMINED
⇒ with V2 = the final volume = 7.88L
P2 = (P1*V1)/V2
P2 = (667*10.0)/7.88
P2 = 846 torr
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Answer:
8.72 × 10^5 moles
Explanation:
To find the number of moles in 5.25 x 10^29 molecules of sucrose, we divide the number of molecules by Avagadro constant (6.02 × 10²³ molecules). That is;
no. of moles = no. of molecules ÷ 6.02 × 10²³ molecules
In this case of sucrose, no of moles contained is as follows;
5.25 × 10^29 ÷ 6.02 × 10²³
5.25/6.02 × 10^ (29-23)
0.872 × 10^6
= 8.72 × 10^5 moles
Answer:
a. ![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. ![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Explanation:
Untuk semua jenis reaksi umum:

Konstanta kesetimbangan ![K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BC%5D%5Ec%20%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Dari pertanyaan yang diberikan:
a. 
Konstanta kesetimbangan:
![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. 
Konstanta kesetimbangan untuk tekanan parsial 
![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Karena Fe3O4 (s) hadir sebagai padatan.