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HACTEHA [7]
3 years ago
14

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

re volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-
Chemistry
1 answer:
My name is Ann [436]3 years ago
6 0

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.

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According to the reaction equation:

and by using ICE table:

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initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
5 0
3 years ago
Plz help (it’s a picture)
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Which of the following statements is true?
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A sample of flammable liquid is placed into an enclosed cylinder which is then fitted with a movable piston. Initially the cylin
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Answer:

12.09 L

Explanation:

Step 1: Convert 826.1 mmHg to atm

We will use the conversion factor 760 mmHg = 1 atm.

826.1 mmHg × 1 atm/760 mmHg = 1.087 atm

Step 2: Convert 427.8 J to L.atm

We will use the conversion factor 101.3 J = 1 L.atm.

427.8 J × 1 L.atm/101.3 J = 4.223 L.atm

Step 3: Calculate the change in the volume

Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:

w = P × ΔV

ΔV = w/P

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5 0
2 years ago
A sample of nitrogen gas is collected over water at temperature of 20.0˚C. What is the pressure of the nitrogen gas if atmospher
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Answer:

P_N=0.987atm

Explanation:

Hello there!

In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

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4 0
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