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3 years ago

Air is made of mostly oxygen, nitrogen, argon, carbon dioxide, and small amounts of other substances. Air is a ___________.

Chemistry

2 answers:

mixas84 [53]3 years ago

5 0

<span>Air is a homogeneous mixture.</span>

Dimas [21]3 years ago

3 0

Im sure its none of them

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Match the compounds as organic (O) or inorganic (I).

SashulF [63]

H2O2(I)
C6H6(O)
CO2(I)
C2H6(O)
HNO3(I)

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2 years ago

Read 2 more answers

What are alternative periodic tables

aksik [14]

Tabulations of chemical elements differing in their organization from the traditional seen periodic system

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2 years ago

The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal

bonufazy [111]

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = $\frac{0.840}{148.2}moles$ of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release $(5539\times 0.00567)kJ$ of heat or 31.41 kJ of heat

6.60 kJ of heat increases $1^{0}\textrm{C}$ temperature of calorimeter.

So, 31.41 kJ of heat increases $(\frac{1}{6.60}\times 31.41)^{0}\textrm{C}$ or $4.76^{0}\textrm{C}$ temperature of calorimeter

So, the final temperature of calorimeter = $(20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}$

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3 years ago

A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o

Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where $\frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$

We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$