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3 years ago

What are electrolytes

1 answer:

3 0

An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. The dissolved electrolyte separates into cations and anions, which disperse uniformly through the solvent. Electrically, such a solution is neutral.

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A C=C bond is usually what kind of bond? Ionic Polar covalent Hydrogen Covalent

HACTEHA [7]

it is a polar covalent bond

7 0

4 years ago

What is the energy released in the fission reaction 1 0 n + 235 92 U → 131 53 I + 89 39 Y + 16 1 0 n 0 1 n + 92 235 U → 53 131 I

mihalych1998 [28]

<u>Answer:</u> The energy released in the given nuclear reaction is 94.99 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{53}^{131}\textrm{I}+_{39}^{89}\textrm{Y}+16_{0}^1\textrm{n}$

We are given:

Mass of $_{92}^{235}\textrm{U}$ = 235.043924 u

Mass of $_{0}^{1}\textrm{n}$ = 1.008665 u

Mass of $_{53}^{131}\textrm{I}$ = 130.9061246 u

Mass of $_{39}^{89}\textrm{Y}$ = 88.9058483 u u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(m_U+m_{n})-(m_{I}+m_{Y}+16m_{n})\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.1019761u)\times c^2$

$E=(0.1019761u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=94.99MeV$

Hence, the energy released in the given nuclear reaction is 94.99 MeV.

3 0

3 years ago

How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(C=0.895 Jig Cy from 22.5°C to 55 0

dsp73

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

$q=mc\Delta T\\\\q=mc(T_2-T_1)$

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum = $0.895J/g^oC$

$\Delta T$ = change in temperature

$T_1$ = initial temperature = $22.5^oC$

$T_2$ = final temperature = $55.0^oC$

Putting values in above equation, we get:

$q=223g\times 0.895J/g^oC\times (55.0-22.5)^oC$

$q=6486.5125J$

Therefore, the heat energy needed would be, 6486.5125 J

5 0

3 years ago

Calculate the concentration of buffer components present in 210.00 mL of a buffer solution that contains 0.300 M NH4Cl and 0.300

Lynna [10]

Answer:

[NH3] = 0.270M

[NH4Cl] = 0.327M

Explanation:

The HNO3 will react with the weak base, NH3, as follows:

HNO₃ + NH₃ → NH₄⁺ + NO₃⁻

Initial moles of each specie of the buffer:

NH3 = NH4⁺ 0.210L * (0.300mol/L) = 0.063moles

The moles added of HNO3 = Additional moles of NH4Cl and the moles substracted of NH3:

0.001L * (6mol / L) = 0.006 moles.

After the addition:

Moles NH3 = 0.063mol - 0.006mol = 0.057moles

Moles NH4Cl = 0.063mol + 0.006mol = 0.0069moles

And their concentrations are:

[NH3] = 0.057moles / 0.211L = 0.270M

[NH4Cl] = 0.069moles / 0.211L = 0.327M

7 0

4 years ago

Under normal conditions, NaOH completely ionizes as: NaOH ->

viktelen [127]

the answer is false because naoh does ionizes

6 0

3 years ago

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