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3 years ago

Which compound can act as both a Brønsted-Lowry acid and a Brønsted-Lowry base?

2 answers:

7 0

Water.
In certain circumstances, water acts as a proton donor.
e.g NH3(aq) + H2O(l) --> NH4+(aq) + OH-(aq)

In other circumstances, water can act as a proton acceptor.
e.g HNO3(aq) + H2O(l) --> H3O+(aq) + NO3-(aq)

denis23 [38]3 years ago

5 0

Answer:

A: water

Explanation:

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Complete the following radioactive decay problem. 222 86 RN to 4 2 HE + ?

Andreas93 [3]

Answer:

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Explanation:

The given nuclear reaction shows alpha decay.

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Properties of alpha radiations:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to parent atom the starting atom.

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₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy

8 0

3 years ago

Which factor is generally responsible for high melting points?

kherson [118]

The factor that is generally responsible for higher melting point is intermolecular forces. The compounds that are covalent in nature are made of molecules rather than ions. It has been seen that some of the covalent compounds have polar molecules at one end, due to which the one end has more electronegative force than the other. The electrostatic force that is bounding the compound is the main cause of higher melting point of this compound. So it is true that with the increase of polarity of a compound creates higher melting point. .. hope I helped

3 0

3 years ago

For the reaction PCl5(g) + heat PCl3(g) + Cl2(g), what will happen when the temperature is decreased

vovikov84 [41]

There will be a shift towards the reactants

5 0

3 years ago

Read 2 more answers

What was a Parthian battery?

timama [110]

Answer:

Scientists believe the batteries (if that is their correct function) were used to electroplate items such as putting a layer of one metal (gold) onto the surface of another (silver), a method still practiced in Iraq today.

Explanation:

8 0

4 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

6 0

3 years ago