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mezya [45]
2 years ago
14

Enter the balanced NET IONIC equation for the potentially unbalanced equation AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq).

Chemistry
1 answer:
Zigmanuir [339]2 years ago
4 0

Answer:

Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)

Explanation:

Hello!

In this case, since the net ionic equations are ionic representations of the molecular equation in which the spectator ions (those at both reactants and products sides) are cancelled out, we first write the complete ionic equation for this reaction, considering that the solid silver chloride is not ionized due to its precipitation:

Ag^+(aq)+NO_3^-(aq)+Na^+(aq)+Cl^-(aq)\rightarrow AgCl(s)+Na^+(aq)+NO_3^-(aq)

Whereas the nitrate and sodium ions are cancelled out for the aforementioned reason as they are the spectator ions, to obtain:

Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)

Which is the required net ionic equation.

Best regards!

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For the reaction 3h2(g) + n2(g) 2nh3(g), kc = 9.0 at 350°c. calculate g° at 350°c.
miss Akunina [59]
When ΔG° is the change in Gibbs free energy

So according to ΔG° formula:

ΔG° =  - R*T*(㏑K)

here when K = [NH3]^2/[N2][H2]^3 = Kc 

and Kc = 9 

and when T is the temperature in Kelvin = 350 + 273 = 623 K

and R is the universal gas constant = 8.314 1/mol.K

So by substitution in ΔG° formula:

∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)

           = - 4536 
4 0
3 years ago
During photosynthesis, the following reaction takes place: carbon dioxide + water + light energy → sugar + oxygen During cellula
Pachacha [2.7K]

Answer:

It cost Evan $17.70 to send 177 text messages. How many text messages did he send if he spent $19.10?

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7 0
2 years ago
in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola
Valentin [98]

Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP  = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula  

                               PV = nRT

Rearrange the equation for Volume

                            V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

                   V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

                   V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

6 0
3 years ago
The [H+] of a solution can be determined from the pH using the equation: [H+] = 10 - pH ... Where: [H+] = hydrogen ion concentra
padilas [110]

Answer:

pH = -log₁₀ [H⁺]

Explanation:

pH is a value in chemistry used in to measure solution trying to determine each quality, purity, risks for health of some products, etc.

As you write in the question, [H⁺] = 10^(-pH)

Using logarithm law (log (m^(p) = p log(m):

log₁₀ [H⁺] = -pH

And

<h3>pH = -log₁₀ [H⁺]</h3>
8 0
3 years ago
A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

  • pH = pKa + log\frac{[CH_3COO^-]}{[CH_3COOH]}

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
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The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
6 0
3 years ago
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