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3 years ago

13

An element has two isotopes. 90% of the isotopes have a mass number of 20 amu, while 10% have a mass number of 22 amu. Calculate

the Average Atomic Mass of the element

1 answer:

jek_recluse [69]3 years ago

8 0

Explanation:

aam= 90%•20amu+10%•22amu/100

aam= 2020/100

aam= 20.2

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Because water molecules are polar and carbon dioxide molecules are nonpolar,

Gala2k [10]

The correct is D.

Water is a polar molecule and it has polar bonds, which carry partially positive and partially negative charges. This polar bond increases the attraction between molecules of water and thus it requires a greater energy to break the bond between the molecules of water compare to carbon dioxide, which is a non polar molecule. Thus, water has a higher boiling point than carbon dioxide.

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Answer:

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2 years ago

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gayaneshka [121]

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Air at 293 K and 750 mm Hg pressure has a relative humidity of 80%. What is its percent humidity? The vapour pressure of water a

Mekhanik [1.2K]

Answer : The correct option is, (c) 79.62

Explanation :

The formula used for percent humidity is:

$\text{Percent humidity}=\text{Relative humidity}\times \frac{p-p^o_v}{p-p_v}\times 100$ ..........(1)

The formula used for relative humidity is:

$\text{Relative humidity}=\frac{p_v}{p^o_v}$ ...........(2)

where,

$p_v$ = partial pressure of water vapor

$p^o_v$ = vapor pressure of water

p = total pressure

First we have to calculate the partial pressure of water vapor by using equation 2.

Given:

$p=750mmHg$

$p^o_v=17.5mmHg$

Relative humidity = 80 % = 0.80

Now put all the given values in equation 2, we get:

$0.80=\frac{p_v}{17.5mmHg}$

$p_v=14mmHg$

Now we have to calculate the percent humidity by using equation 1.

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3 years ago

2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??

Zepler [3.9K]

Answer:

$d=1.78\times 10^{-7}\ Dg/\mu m^3$

Explanation:

Given,

The density of Helium is $1.78\times 10^4\ g/cm^3$

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

$d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3$

So, the density of Helium is equal to $1.78\times 10^{-7}\ Dg/\mu m^3$ .

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2 years ago