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PSYCHO15rus [73]
2 years ago
12

Answer the following questions…

Chemistry
1 answer:
Zina [86]2 years ago
3 0

The concentration of the hydronium ion in hydrochloric acid is 0.0045 M, and the pH of the solution is 2.34.

<h3>What is pH?</h3>

pH is the potential of the hydrogen or the hydronium ions in the aqueous solution.

As the solution contains 4.5 \times 10^{-3} \;\rm  M\; HCl the concentration of the hydronium ion will be the same, 4.5 \times 10^{-3} \;\rm  M.

The pH of the solution is calculated as:

\begin{aligned} \rm pH &= \rm -log[H^{+}]\\\\&= - \rm log (4.5 \times 10^{-3})\\\\&= 2.34\end{aligned}

The concentration of the hydroxide ion is calculated from pH and hydronium ion as:

\begin{aligned} \rm [H_{3}O^{+}][OH^{-}] &= 10^{-14}\\\\&= \dfrac{1 \times 10^{-14}}{4.5 \times 10^{-3}}\\\\&= 2.2 \times 10^{12}\end{aligned}

Now, for the calcium hydroxide solution, the calculations are shown as,

\begin{aligned} \rm (H_3}\rm O^{+}) &= \rm antilog (-pH)\\\\&= \rm antilog (-8)\\\\&= 10^{-8} \;\rm M\end{aligned}

pOH is calculated as:

\begin{aligned} \rm pOH &= 14- 8 = 6\\\\\rm [OH^{-}] &=  \rm antilog (-6)\\\\&= 10^{-6} \end{aligned}

The concentration of calcium hydroxide is calculated as:

\begin{aligned} &= \dfrac{1}{2} \times \rm [OH^{-}]\\\\&= 5 \times 10^{-4} \;\rm M\end{aligned}

Therefore, the pH and the pOH give the concentration of the hydrogen or the hydronium ion and the hydroxide ion.

Learn more about pH and pOH here:

brainly.com/question/16062632

#SPJ1

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What is mass in g of 15.5 moles of calcium?
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Answer:

621.2090000000001 grams

Explanation:

1 moles Calcium to grams = 40.078 grams

15.5*40.078 = 621.2090000000001 g

4 0
3 years ago
Zinc metal (Zn) reacts with hydrochloric acid (HCI) to produce hydrogen gas (H) and zinc chloride (ZnCl2). A scientist has powde
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2 years ago
A buffer is prepared by adding 139 mL of 0.39 M NH3 to 169 mL of 0.19 M NH4NO3. What is the pH of the final solution? (Assume th
Paha777 [63]

Answer:

pH = 9.48

Explanation:

We have first to realize that NH₃ is a weak base:

NH₃ + H₂O ⇔ NH₄⁺ + OH⁻     Kb = 1.8 x 10⁻⁵

and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.

Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:

pOH = pKb + log ( [ conjugate acid ] / [  weak base ]

mol NH₃ = 0.139 L x 0.39 M = 0.054 mol

mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol

Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)

pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52

pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48

The solution is basic which agrees  with NH₃ being  a weak base.

5 0
3 years ago
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
The reaction 2NO(g)+Cl2(g)→2NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate o
Luba_88 [7]

Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.

Explanation : Given,

\frac{d[NO]}{dt} =21 torr/min

The balanced chemical reaction is,

2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)

The rate of disappearance of NO = -\frac{1}{2}\frac{d[NO]}{dt}

The rate of disappearance of Cl_2 = -\frac{d[Cl_2]}{dt}

The rate of formation of NOCl = \frac{1}{2}\frac{d[NOCl]}{dt}

As we know that,

\frac{d[NO]}{dt} =21 torr/min

So,

-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}

\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min

And,

\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}

\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min

Now we have to calculate the rate change.

Rate change = Reactant rate - Product rate

Rate change = (21 + 10.5) - 21 = 10.5 torr/min

Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.

8 0
3 years ago
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