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asambeis [7]
3 years ago
12

In this reaction what is the correct coefficient for sodium chloride? pb(no3)2 ? nacl → pbcl2 ? nano3

Chemistry
2 answers:
weqwewe [10]3 years ago
8 0

Answer:

2

Explanation:

The unbalanced reaction is shown below as:-

Pb(NO_3)_2+NaCl\rightarrow PbCl_2+NaNO_3

On the left hand side,  

There is 1 lead atom and 2 nitrogen atoms and 6 oxygen atoms  and 1 sodium atom and 1 chlorine atom

On the right hand side,  

There is 1 lead atom and 2 chlorine atoms and 3 oxygen atoms  and 1 sodium atom

Thus,  

Right side, NaNO_3 must be multiplied by 2 so to balance nitrogen and oxygen.

Left side, NaCl is multiplied by 2 so to balance the whole reaction.

Thus, the balanced reaction is:-

Pb(NO_3)_2+2NaCl\rightarrow PbCl_2+2NaNO_3

<u>Coefficient of the sodium chloride in the balanced equation - 2</u>

Dovator [93]3 years ago
4 0
<span>pb(no3)2 + 2 nacl → pbcl2 + 2nano3</span>
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Answer:

Arterioles

Explanation:

Arterioles

5 0
3 years ago
Read 2 more answers
What is the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution?
RoseWind [281]

Answer: 0.4M

Explanation:

Given that,

Amount of moles of NaOH (n) = ?

Mass of NaOH in grams = 40.0g

For molar mass of NaOH, use the atomic masses: Na = 23g; O = 16g; H = 1g

NaOH = (23g + 16g + 1g)

= 40g/mol

Since, n = mass in grams / molar mass

n = 40.0g / 40.0g/mol

n = 1 mole

Volume of NaOH solution (v) = 2.5 L

Concentration of NaOH solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

c = 1 mole / 2.5 L

c = 0.4 mol/L (Concentration in mol/L is the same as Molarity, M)

Thus, the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution is 0.4 mol/L or 0.4M

7 0
3 years ago
If 1.00L of muriatic acid w/ a pH of 2.5 is poured to 8.00L water, what is the new molar concentration of the muriatic solution?
Daniel [21]

Answer:

The answer to your question is:  C2 = 0.0004 M

Explanation:

Data

pH = 2.5; V = 1.0 L

V2 = 8.0 L  C2 = ?

Formula

C1V1 = C2V2

C2 = C1V1 / V2

pH = -log[H⁺]

Process

[H⁺] = antilog -pH

[H⁺] = antilog (-2.5)

[H⁺] = 0.003 M = C1

Finally

                          (0.003)(1 l) = C2(8)

                           C2 = 0.003 / 8

                           C2 = 0.0004 M

8 0
3 years ago
What is acid sulfuric's function ?​
masya89 [10]

Answer:

The answer is below

Explanation:

Sulfuric acid is used for various purposes, some of which include the following:

1. It is used in the production of various manufactured goods.

2. It is used in the manufacturing of chemicals

3. It is also used in the making of fertilizer

4. It is used in the refining process of petroleum products

5. It is used in the processing of metals

3 0
3 years ago
Calculate the enthalpy of the reaction
harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)     \Delta H_A=+2035kJ

(2) 2B(s)+3H_2(g)\rightarrow B_2H_6(g)    \Delta H_B=+36kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_C=-285kJ

(4) H_2O(l)\rightarrow H_2O(g)    \Delta H_D=+44kJ

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D

\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)

\Delta H=-2552kJ

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0
3 years ago
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