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3 years ago

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Rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3) in order o

f decreasing concentration of CO2.
~
I need to know how to physically find the percentages of CO2 in H2O2 & NaHCO3

1 answer:

Komok [63]3 years ago

6 0

Astronomers use chemical signatures to determine<span> the age and .... </span>Gas<span> mixtures that contain more than </span>4<span>% hydrogen in </span>air<span> are potentially explosive. ... water vapor, </span>carbon dioxide<span>, and several other </span><span>gases</span>

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david’s mom wants to calculate how much it will cost to drive from los angeles, ca, to san francisco, ca. gas costs $4.00 a gall

grin007 [14]

She needs to know the distance from LA to SF.

Call that distance x. Then you can calculate the cost from:

# of gallons consumed: x miles / 38 miles/ gallon = (x/38) gallons

After that,

cost = # gallons * cost per gallon = (x/38) gallons * 4 $/gallon = 4x/38 $

3 0

3 years ago

Read 2 more answers

In the lab there is a vile of mercury and it has a mass of 43.0 grams. If the density of mercury is 13.6 g/mL, what is the volum

Vitek1552 [10]

Answer:

<h2>The answer is 3.16 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density}$

From the question

mass = 43 g

density = 13.6 g/mL

The volume is

$volume = \frac{43}{13.6} \\ = 3.161764$

We have the final answer as

<h3>3.16 mL</h3>

Hope this helps you

8 0

3 years ago

The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio

iren2701 [21]

Answer : The correct option is, (B) $CO_2$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of unknown gas = $11.9\text{ mL }min^{-1}$

$R_2$ = rate of effusion of oxygen gas = $14.0\text{ mL }min^{-1}$

$M_1$ = molar mass of unknown gas = ?

$M_2$ = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}$

$M_1=44.2g/mole$

The unknown gas could be carbon dioxide $(CO_2)$ that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide $(CO_2)$

5 0

3 years ago

How many moles of chromium metal equal 7.52 × 1022 atoms Cr?

grigory [225]

Answer:

0.125 mole

Explanation:

From Avogadro's hypothesis, we discovered that 1 mole of any substance contains 6.02x10^23 atoms.

The statement above suggests that 1 mole of chromium contains 6.02x10^23 atoms.

Now, if 1mole of of chromium contains 6.02x10^23 atoms,

Then Xmol of chromium contains 7.52x10^22 atoms i.e

Xmol of chromium = 7.52x10^22/6.02x10^23

Xmol of chromium = 0.125 mole

Therefore, 0.125 mole of chromium contains 7.52x10^22 atoms

8 0

3 years ago

Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei

Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is $R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length <em>a </em>and atomic radius <em>R </em>are related through

$a=\frac{4R}{\sqrt{3} }$

So the volume of the unit cell $V_{c}$ is

$V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}$

We can compute the theoretical density ρ through the following relationship

$\rho=\frac{nA}{V_{c}N_{a}}$

where

n = number of atoms associated with each unit cell

A = atomic weight

$V_{c}$ = volume of the unit cell

$N_{a}$ = Avogadro’s number ( $6.023 \times 10^{23}$ atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio <em>R</em> as follows:

$\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}$

Solving for <em>R</em>

$\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}$

Substitution for the various parameters into above equation yields

$R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

7 0

3 years ago