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Klio2033 [76]
3 years ago
13

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

ionic equation for this reaction by filling in the blanks. The phase symbols and charges on species are already provided.
Complete and balance the net ionic equation for this reaction.
Chemistry
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

See explanation

Explanation:

The molecular equation shows all the compounds involved in the reaction.

The molecular equation is as follows;

2NaF(aq) + Pb(NO3)2(aq) -------> PbF2(s) + 2NaNO3(aq)

The complete ionic equation shows all the ions involved in the reaction

The complete ionic equation;

2Na^+(aq) + 2F^-(aq) + Pb^2+(aq) + 2NO3^-(aq) -------->PbF(s) + 2Na^+(aq) +2NO3^-(aq)

The net Ionic equation shows the ions that actually participated in the reaction

The net ionic equation is;

2F^-(aq) + Pb^2+(aq)--------> PbF(s)

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In a mixture of argon and hydrogen, occupying a volume of 1.18 l at 894.6 mmhg and 44.1oc, it is found that the total mass of th
alexandr1967 [171]

To solve the question, there is a need to use the equation:

PV = nRT

(894.6/760) × 1.18 = n × 0.0821 × (273 + 44.1)

By solving we get:

Total moles, n = 0.053

Assume, the moles of argon as a and of hydrogen as b,

So,

40 × a + 2 × b = 1.25 --------- (i)

a + b = 0.053 ------- (ii)

By solving i and ii we get:

a = 0.03,

Thus, mole fraction of Ar = XAr = 0.03/0.053 = 0.57

So,

Partial pressure of Ar = 894.6 × XAr = 894.6 × 0.57 = 509.92 mm Hg



6 0
3 years ago
How many moles of carbon in 130g of C2H6
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 7.22 moles of C2H6. Since there are 2 carbon atoms per C2H6, we must multiply the number of moles of C2H6 by 2 to get the number of moles of Carbon which is 14.4 or 14 if using two sig figs.
5 0
4 years ago
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Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing tin
mojhsa [17]
  The  half  reactions    as  they  occur   at  each  electrode

is  as  follows

at  the  anode  Sn(s)  =sn^2+(aq)  +  2e  -
at  the  cathode  2  ag^+(aq)   + 2e -  =  2Ag (s)

net  cell  reaction  =   Sn (s)  +  2Ag^+(aq)  =  sn^2+ (aq) +  2  Ag  (s)
8 0
3 years ago
Determine the molar mass of Ze(NO3)2. (Ze=55.85 N=14.01 O=16.00) *<br> Help
Arturiano [62]

Answer:

179.87 g/mol

Explanation:

First you need to determine the number of each elements in the molecule.  This information comes from the molecular formula.  

Ze(NO3)2 tells us that there is 1 Ze atom and 2 NO3 anions per molecule.  each NO3 anion will have 1 nitrogen and 3 oxygens.  Due to that, one molecule of Ze(NO3)2 will have 1 atom of Ze, 2 atoms of nitrogen (N), and 6 atoms of oxygen (O).

Next you need to add all of the individual atom's molar masses to get the over all molar masses.  The molar masses of each element is in the question but it can also be found on the periodic table.

molar mass of Ze(NO3)2 = 55.85g/mol + (14.01g/mol*2) + (16.00g/mol*6)

molar mass of Ze(NO3)2 = 179.87 g/mol

I hope this helps.

5 0
2 years ago
Cierto elemento presenta dos isótopos cuyos números de masa suman 110 y cuyos números de neutrones suman 58 ¿Cuál es el número a
Anettt [7]

Answer:

El número atómico de cada uno de los átomos es 26

Explanation:

El número de masa es la suma de las masas del protón y el neutrón de un átomo.

El número atómico es el número de protones en el átomo.

Los parámetros dados son;

La suma del número másico de ambos átomos = 110

La suma de los neutrones = 58

Por lo tanto, sea el número de protones y neutrones en un isótopo = P₁ y N₁ y el número de protones y neutrones en el otro isótopo = P₂ y N₂

Tenemos;

 P₁ + N₁ + P₂ + N₂ = 110

N₁ + N₂ = 58

Por lo tanto;

P₁ + P₂ = 110 - (N₁ + N₂)

P₁ + P₂ = 110 - 58 = 52

Dado que los isótopos son del mismo elemento, sus protones serán iguales, por lo tanto;

P₁ = P₂

P₁ + P₂ = P₁ + P₁ = 2 × P₁

P₁ + P₂ = 52

2 × P₁ = 52

P₁ = 52/2 = 26 = P₂

El número atómico de ambos átomos es el número de protones en el átomo que es 26.

El número atómico del elemento del átomo es 26

6 0
3 years ago
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