All categories

3 years ago

A picture of a patient’s lung showed it to be shriveled and darker than usual.

Chemistry

2 answers:

Strike441 [17]3 years ago

3 0

Answer:

Smoking

Explanation:

It has been proven time and time again to ruin lungs, effectively shortenning ones lifespan.

suter [353]3 years ago

3 0

Explanation:

Smoking, because smoking causes these types of symptoms in long time smokers and people who start to smoke. also I know the answer cause my dad's a nurse.

You might be interested in

The sp of pbbr2 is 6. 60×10−6. what is the molar solubility of pbbr2 in pure water?

Inessa05 [86]

Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.

Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.

Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.

Let the molar solubility be S upon dissociation.

PbBr₂ or Lead Bromide dissociates in pure water as follows:

PbBr₂ ----------> Pb⁺² + Br⁻

S 2S

Ksp = [Pb⁺²] [ Br⁻]

Ksp = (S) (2S)²

Ksp = 4S³

6.60 × 10⁻⁶ = 4S³

S = 0.0118M

Hence, the Molar solubility S is 0.0118M.

Learn more about Molar solubility here, brainly.com/question/16243859

#SPJ4

6 0

2 years ago

2) Examine the molecules below.<br> Circle the molecules that can also be classified as compounds.

sergejj [24]

Answer:

3rd

Explanation:

its the way they look

4 0

2 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

3 years ago

How many atoms in 7.55 mole of Iron atom

vagabundo [1.1K]

7.55 x 6.02 x 10²³ = 4.55 x 10²⁴ atoms

4 0

2 years ago

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .

8 0

3 years ago