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nalin [4]
2 years ago
7

Calculate the maximum concentration (in m) of silver ions (ag+) in a solution that contains 0.025 m of co32-. the ksp of ag2co3

is 8.1 x 10-12.
Chemistry
1 answer:
Helen [10]2 years ago
5 0
Equilibrium equation is

<span>Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) </span>

<span>From the reaction equation above, the formula for Ksp: </span>

<span>Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 </span>
<span>You know  [CO32-], so you can solve for [Ag+] as: </span>
<span>(8.1 x 10^-12) = [Ag+]^2 (0.025) </span>
<span>[Ag+]^2 = 3.24 x 10^-10 </span>
<span>[Ag+] = 1.8 x 10^-5 M </span>
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A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Lit
My name is Ann [436]

Answer:

M=0.213M

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-

n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-

n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-

Finally, we compute the molarity:

M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M

Regards.

6 0
2 years ago
What is the absolute structural necessity for an alcohol to be oxidized with chromium trioxide?
4vir4ik [10]
The alcohol being oxidized must not be a tertiary alcohol.

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2 years ago
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2.00 liters of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated until a volume of 20.0 liters and a pressure
dedylja [7]

Answer:  The new temperature is 10643 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 750.0 mm Hg = 0.98 atm   (760mmHg=1atm)

P_2 = final pressure of gas = 3.50 atm

V_1 = initial volume of gas = 2.00 L

V_2 = final volume of gas = 20.0 L

T_1 = initial temperature of gas = 25.0^oC=273+25.0=298.0K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{0.98\times 2.00}{298.0K}=\frac{3.50\times 20.0}{T_2}

T_2=10643K

Thus the new temperature is 10643 K

6 0
2 years ago
A compound composed of 3.3% H, 19.3% C, and 77.4% O3.3% H, 19.3% C, and 77.4% O has a molar mass of approximately 60 g/mol.. Wha
11111nata11111 [884]

Answer:

Molecular formula of the compound = H₂CO₃

Explanation:

First, the empirical formula of the compound is determined

Percentage by mass of each element is given as shown below:

H = 3.3% ; C = 19.9%; O = 77.4%

Mole ratio of the elements= percentage mass/ molar mass

H = 3.3/ 1 = 3.3

C = 19.3/12 = 1.6

O = 77.4/16 = 4.8

whole number ratio is obtained by dividing through with the smallest ratio

H = 3.3/1.6; C = 1.6/1.6; O = 4.8/1.6

H : C : O = 2 : 1 : 3

Empirical formula = H₂CO₃

Molecular formula/mass = n(empirical formula/mass)

60 = n(2*1 + 12*1 + 16*3)

60 = n(62)

n = 60/62 = 0.96

n is approximately = 1

Therefore, molecular formula of the compound = (H₂CO₃) * 1

Molecular formula of the compound = H₂CO₃

6 0
2 years ago
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