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3 years ago

Both "Ode to the West wind" and "Ode for Melancholy"

1 answer:

5 0

The answer that would best complete the given statement above would be option A. Both "Ode to the West wind" and "Ode for Melancholy" praise <span>something non-human. Hope this answers your question. Have a great day ahead!</span>

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Number of moles in 31.99g O2 <br> And 176.8g NaCl

sergey [27]

#O_2

$\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longmapsto No\:of\:moles=\dfrac{31.99}{32}$

$\\ \sf\longmapsto No\:of\:moles\approx 1$

#NaCl

$\\ \sf\longmapsto No\:of\:moles=\dfrac{176.8}{58.5}$

$\\ \sf\longmapsto No\;of\:moles=3mol$

8 0

2 years ago

What is the identity of the element whose isotopes you have selected

Pachacha [2.7K]

Answer:

Rubidium

Explanation:

Nil

5 0

2 years ago

NaF + HCL=> NACL + HF

Jobisdone [24]

Answer:

NaF + HCl --> NaCl + HF

Explanation:

hope it helps you

4 0

2 years ago

Felicity lists the factors that affect the shape of Florida

umka2103 [35]

1. Humidity

2. Temperature

3. Rainfall

In general, the electron pairs in a molecule that will adopt a shape to maximize their distance far apart, even charges repel each other as well as unlike charges attract,

for example H2O has two lone electrons pairs but it adopts an angular shape. Hope this help!!

5 0

3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$