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4 years ago

Identify ammonia.

1 answer:

8 0

The answer is D. Ammonia is a weak electrolyte and a weak base. Ammonia produces ions when dissolving but remains predominantly as molecules that are not ionized. This is what makes it a weak electrolyte. Ammonia also does not fully ionize in a solution. This is what makes is a weak base.

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Water supplies are often treated with chlorine as one of the processing steps in treating wastewater. Estimate the liquid diffus

jekas [21]

Answer:

⇒D_AB= 1.21×10^(-9)

Explanation:

Wike chang equation is given as:

$D_{AB}= \frac{117.3\times10^{-18}\times\(\phi\times M_B)^{0.5}\times T}{\mu\times\nu^{0.6}}$

Where

D_AB= diffusivity of chlorine in water

Φ= 2.26 for water as solvent

ν= 0.0484 for chlorine as solute

M_B = Molecular weight of water

τ= temperature=289 K

μ= viscosity = 1.1×10^{-3}

Now putting these values in the above equation we get

$D_{AB}= \frac{117.3\times10^{-18}\times\(\2.26\times18)^{0.5}\times289}{\1.1\times10^{-3}\times\0.0484^{0.6}}$

⇒D_AB= 1.21×10^(-9)

7 0

3 years ago

Read 2 more answers

Help ;-; Will give Brainliest and 25 Points

Yuri [45]

I don’t see what you need help with but thanks:)

3 0

3 years ago

Read 2 more answers

Make observations. describe the system in each image by filling out the table below.

Snezhnost [94]

Answer:

ok??????????????

Explanation:

5 0

3 years ago

What atoms are likely to lose electrons to form cations

Anika [276]

Hey Jose! Thanks for asking a question here on Brainly.

⚪ Potassium
⚪ Magnesium
⚪ Sodium (aka Na)
⚪ Lithium (aka Li)

These are all atoms that are likely to lose electrons in order to form cations. Hope this helps. -UF aka Nadia

5 0

3 years ago

Read 2 more answers

The natural distribution of the isotopes of a hypothetical element is 60.795% at a mass of 281.99481 u, 22.122% at a mass of 283

fenix001 [56]

Answer: The average atomic mass of the element is 283.291 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 281.99481 amu

Percentage abundance of isotope 1 = 60.795 %

Fractional abundance of isotope 1 = 0.60795

For isotope 2:

Mass of isotope 2 = 283.99570 amu

Percentage abundance of isotope 2 = 22.122 %

Fractional abundance of isotope 2 = 0.22122

For isotope 3:

Mass of isotope 3 = 286.99423 amu

Percentage abundance of isotope 3 = [100 - (60.795 + 22.122)] = 17.083 %

Fractional abundance of isotope 1 = 0.17083

Putting values in equation 1, we get:

$\text{Average atomic mass of element}=[(281.99481\times 0.60795)+(283.99570\times 0.22122)+(286.99423\times 0.17083)]\\\\\text{Average atomic mass of element}=283.291amu$

Hence, the average atomic mass of the element is 283.291 amu

5 0

4 years ago