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3 years ago

What is the main difference between protons and neutrons?

Chemistry

2 answers:

kykrilka [37]3 years ago

3 0

Answer:

Protons have a much larger volume than neutrons.

Marrrta [24]3 years ago

3 0

Protons are positively charged +1 while neutrons charge = 0

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IrinaVladis [17]

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the second stage of cell division. its between prophase and anaphase during the chromosomes become attached to spindle fibers.

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3 years ago

. When a newspaper is left in direct sunlight for a few days, the paper begins to turn yellow. The yellow

34kurt

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It is being changed by the sun

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2 years ago

Select all of the answers that apply. Which of the following are ways the atmosphere supports life?

kupik [55]

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the atmosphere supports life by giving us simple things like wood.

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3 years ago

1. What mass of Mgo will I make from 48 g of Mg?<br> 2Mg + O2 + 2MgO

statuscvo [17]

Answer:

80.6 g

Explanation:

MgO=24.3+16=40.3 g

48.6/24.3= 2 moles

64/2x16= 2 moles

2x40.3= 80.6g

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2 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

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3 years ago