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2 years ago

Classify each cation as a weak acid or pH neutral (neither acidic nor basic).

Chemistry

1 answer:

Alexeev081 [22]2 years ago

3 0

Answer:

a. Na+ is pH neutral

b. Ni2+ = weak acid

c. NH4+ = Weak acid

Explanation:

To know the nature of the cation we need to find the nature of its conjugate base.

If the conjugate base of the ion is a strong base, the ion is neutral.

If the conjugate base is a weak base, the ion is a weak acid:

a. Conjugate base Na+ = NaOH

Sodium hydroxide is a strong base:

Na+ is pH neutral

b. Conjugate base Ni²⁺: Ni(OH)2 is a weak base because is not completely soluble in water. That means:

Ni2+ = weak acid

c. Conjugate base NH4+: NH4OH. Weak base:

NH4+ = Weak acid

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Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The

love history [14]

Answer:

(a) $V_B=11.68L$

(b) $x_{He}=0.533$

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

$n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol$

Thus, since the final pressure is 3.60 bar, we can write:

$P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar$

The moles of helium could be computed via solver as:

$n_{He}=2.358mol$

Or algebraically:

$3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol$

In such a way, the volume of the compartment B is:

$V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\ \\V_B=11.68L$

Finally, he mole fraction of He is:

$x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533$

Regards.

8 0

2 years ago

How many moles of oxygen would be needed to produce 84 moles of sulfur trioxide according to the following balanced chemical equ

olganol [36]

Answer:

126 moles

Explanation:

2S +3 o2=2so3

So if 2 moles of so3 required 3 moles of oxygen

. So 84 moles of so3 will require 84*3/2=126 moles of oxygen

4 0

3 years ago

SHOW WORK AND INCLUDE UNITS

Ber [7]

Answer:

36365.4 Joules

Explanation:

The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C), and change in temperature (Φ)

Thus, Q = MCΦ

Since, M = 45.4 g

C = 3.56 J/g°C,

Φ = 250°C - 25°C = 225°C

Q = 45.4g x 3.56J/g°C x 225°C

Q= 36365.4 Joules

Thus, 36365.4 Joules of heat energy is released when the lithium is cooled.

6 0

2 years ago

For the reaction: Cr2O7^2- + NH4^+ → Cr2O3 + N2, which is the oxidizing agent?

lord [1]

Answer:

(NH4)2Cr2O7

Explanation:

Hope this somehow helped.

3 0

2 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago