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3 years ago

Suggest why helium can be used as a carrier gas but oxygen cannot.

Chemistry

2 answers:

SCORPION-xisa [38]3 years ago

7 0

Because helium is lighter then the oxygen .and due to this helium gass carrier go upward easily .

Inga [223]3 years ago

6 0

Helium is lighter then the oxygen and due to this helium gas carrier go upward easily .

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Which of the following is generally FALSE? Group of answer choices Elements in the same group have the same valence electron con

Whitepunk [10]

Answer: Option (c) is the correct answer.

Explanation:

Generally, when we move from top to bottom in a group then there occur increase in the number of electrons due to which there will also occur increase in the number of shells.

As atomic size is the distance between the nucleus and valence shell of an atom. Hence, more is the number of shells present in an atom more will be its atomic radius.

Thus, we can conclude that the statement atomic radius of elements in a group decreases as you go from top to bottom, is generally FALSE.

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3 years ago

Microwave warming Popcorn is a example of

Viefleur [7K]

Answer:

Explanation:

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3 years ago

2c4H10 +13O2 —>8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10

Elena-2011 [213]

Answer:

$\large\boxed{\text{28 mol of O$_{2}$}}$

Explanation:

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol: 4.3

13 mol of O₂ react with 2 mol of 2C₄H₁₀

$\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}$

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3 years ago

I need to know what the answer to number 4 letter a.

QveST [7]

I believe it is C. I hope

4 0

3 years ago

Read 2 more answers

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

3 years ago

Read 2 more answers