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valentinak56 [21]

3 years ago

14

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

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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What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL

Rasek [7]

Density is weight by volume.

First. If you divide the weight by density you can find the volume

Second you must convert the ML in to Liters.

$\frac{321 \frac{g}{1}}{0.84 \frac{g}{mL}} = \frac{321(g)(mL)}{0.84g}=\frac{321mL}{0.84}=382.14mL$

1L=1000ml

$\frac{1L}{1000mL}$

$(382.14mL)( \frac{1L}{1000mL})= \frac{(382.14mL)(L)}{1000mL} =\frac{382.14L}{1000}=0.38214L$

0.38214 Liters.

4 0

3 years ago

What is the molarity of formaldehyde in a solution containing 0.25 grams of formaldehyde per mL?

Vikentia [17]

Answer:

8.33mol/L

Explanation:

First, let us calculate the molar mass of of formaldehyde (CH2O). This is illustrated below:

Molar Mass of CH2O = 12 + (2x1) + 16 = 12 + 2 + 16 = 30g/mol

Mass of CH2O from the question = 0.25g

Number of mole CH2O =?

Number of mole = Mass /Molar Mass

Number of mole of CH2O = 0.25/30 = 8.33x10^-3mole

Now we can calculate the molarity of formaldehyde (CH2O) as follow:

Number of mole of CH2O = 8.33x10^-3mole

Volume = 1mL

Converting 1mL to L, we have:

1000mL = 1L

Therefore 1mL = 1/1000 = 1x10^-3L

Molarity =?

Molarity = mole /Volume

Molarity = 8.33x10^-3mole/1x10^-3L

Molarity = 8.33mol/L

Therefore, the molarity of formaldehyde (CH2O) is 8.33mol/L

3 0

3 years ago

Si2C19<br> What is this compound

lutik1710 [3]

Answer:

BROMOTRIPHENYLSILANE

Explanation:

<h2>MARK ME AS BRAINLIST</h2>

PLZ FOLLOW ME

7 0

2 years ago

When 150. g zinc sulfide are burned in excess oxygen, 68.5 g of zinc oxide are actually produced, along with sulfur dioxide. Det

Bumek [7]

%yield = 54.6%

<h3>Further explanation</h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

(theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

<h3 />

Reaction

2ZnS+3O₂ ⇒ 2ZnO+2SO₂

MW ZnS = 97.474 g/mol

mol ZnS

$\tt \dfrac{150}{97.474}=1.54$

MW ZnO = 81.38 g/mol

mol ZnO (from mol ZnS as limiting reactant, O₂ excess)

$\tt \dfrac{2}{2}\times 1.54=1.54$

Actual ZnO produced

$\tt 1.54\times 81.38=125.33~g$

Theoretical production = 125.388

%yield

$\tt \dfrac{68.5}{125.33}\times 100\%=\boxed{\bold{54.6\%}}$

4 0

3 years ago

Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) => CH 3OH(l) Using the following information: C(graphite) +

Alika [10]

Answer:

$\Delta H$ for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

$C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)$ $\Delta H_{1}= -393.5 kJ$

$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)$ $\Delta H_{2}= (2\times -285.8)kJ$

$CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)$ $\Delta H_{3}= 726.4 kJ$

---------------------------------------------------------------------------------------------------

$C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)$

$\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ$

4 0

3 years ago