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iren [92.7K]
2 years ago
15

What reaction will take place if h2o is added to a mixture of nanh2/nh3? draw the products of the reaction (without sodium ion)?

Chemistry
1 answer:
Maru [420]2 years ago
8 0

When sodium amide i.e.NaNH_{2} reacts with water i.e. H_{2}O results in the formation of sodium hydroxide i.e. NaOH and ammonia NH_{3}.

The chemical reaction is given by:

NaNH_{2}+H_{2}O\rightarrow NH_{3}+NaOH

Now, when ammonia i.e.NH_{3} reacts with water results in the formation of ammonium hydroxide i.e. NH_{4}OH

The chemical reaction is given by:

NH_{3}+H_{2}O\rightarrow NH_4OH

Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).

The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.


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A student titrates 10.00 milliliters of hydrochloric acid of unknown molarity with 1.000 m naoh. it takes 21.17 milliliters of b
Dima020 [189]
Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
 molarity of HCl=0.02117*10/1000
                         =2.117M
4 0
2 years ago
If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
fredd [130]

Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0
2 years ago
Barry Bonds swings a bat which has a mass of 1.5 kg at a velocity of 55 m/s. How many joules of kinetic energy could he give to
larisa86 [58]
KE = mv2
2
KE = ? J
m = 1.5 kg
v = 55 m/s
KE = 1.5 kg x (55 m/s)2
2
KE = 2,268.75 J
8 0
2 years ago
Read 2 more answers
What is the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide? hints
aleksklad [387]
The answer:

<span>the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide:
</span>CaBr2 + AgNO3-----------> AgBr2 + CaNO3, so among the given choices, the
true answe is 
<span>agno3(aq)+cabr2(aq) (as reactants)</span>
4 0
2 years ago
Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp
weeeeeb [17]

Answer : The grams of N_2 consumed is, 89.6 grams.

Solution : Given,

Mass of CaCN_2 = 265 g

Molar mass of CaCN_2 = 80 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate the moles of CaCN_2.

\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles

The given balanced reaction is,

CaC_2+N_2\rightarrow CaCN_2+C

from the reaction, we conclude that

As, 1 mole of CaCN_2 produces from 1 mole of N_2

So, 3.2 moles of CaCN_2 produces from 3.2 moles of N_2

Now we have to calculate the mass of N_2

\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2

\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g

Therefore, the grams of N_2 consumed is, 89.6 grams.

5 0
2 years ago
Read 2 more answers
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