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1 year ago

The mineral galena is composed of lead(II) sulfide and has an average density of 7.46 g/cm³.

1 answer:

7 0

The mineral galena is composed of lead(II) sulfide and has an average density of 7.46 g/cm³. Moles of lead(II) sulfide in 1.00 ft³ of galena are 883 mol PbS.

<h3>What is moles of compound?</h3>

A very large number of molecules, atoms, or other particles are referred to as a "mole," which is a unit of measurement in chemistry. The number of moles in one unit is 6.02214 x 1023, and it is known as Avogadro's Number. These figures are crucial in providing information on the quantity of the constituent elements. A mole of a substance might be as little as a few grams or as much as hundreds of grams.

One mole (or formula unit) of a chemical is equal to 6.022 x 10²³ molecules (ionic). A chemical's molar mass represents the mass of 1 mole of that substance. To put it another way, it tells you how many grams there are in a chemical per mole.

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A 34.4 L sample of oxygen gas at 229°C and 752 torr is cooled to 34°C at 668 torr. The volume of the sample is now

kenny6666 [7]

Answer:

23.55 L

Explanation:

USe the following 'identity' of gs laws

P1 V 1 / T1 = P2 V2 / T2 ( T must be in Kelvin)

re arrange to

P1 V 1 T2 / (T1 P2) = V2 NOW SUB IN THE VALUES

752 * 34.2 * ( 34 + 273.15) / [( 229 + 273.15) * 668] = V2 = 23.55 L

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1 year ago

1. The gradual wearing away or breaking down of rocks by abrasion is a type of

kiruha [24]

Differential weathering or erosion

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3 years ago

Enter your answer in the provided box. The usefulness of radiocarbon dating is limited to objects no older than 50,000 yr. What

guapka [62]

Answer : The percent of the carbon−14 left is, 0.242 %

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

To calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5.73\times 10^3\text{ years}}$

$k=1.205\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the amount left.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant = $1.205\times 10^{-4}\text{ years}^{-1}$

t = time taken for decay process = 50000 years

a = initial amount or moles of the reactant = 7 g

a - x = amount or moles left after decay process = ?

Putting values in above equation, we get:

$1.205\times 10^{-4}=\frac{2.303}{50000\text{ years}}\log\frac{7g}{a-x}$

$a-x=0.0169g$

The amount left of carbon-14 = 0.0169 g

Now we have to calculate the percent of the carbon−14 left.

$\text{Percent of carbon}-14\text{ left}=\frac{\text{Amount left of carbon}-14}{\text{Original amount of carbon}-14}\times 100$

$\text{Percent of carbon}-14\text{ left}=\frac{0.0169g}{7g}\times 100=0.242\%$

Therefore, the percent of the carbon−14 left is, 0.242 %

8 0

3 years ago

3. What determines the frequency (color) of photons?

ivanzaharov [21]

Whatever the photon doesn’t absorb (color)

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3 years ago

(r)-2-butanol reacts with phosphorus tribromide to give a (c4h9br). treatment of a with sodium cyanide in dmf gives b (c5h9)n. b

Ierofanga [76]

The reaction described above is a series of sn2 substitutions. The initial (<em>R</em>)-2-butanol reacts with PBr₃ in pyridine to turn the alcohol functionality into a leaving group as it attaches to the phosphorus of PBr₃. One of the bromides from PBr₃ eventually displaces the oxygen atom with an sn2 substitution. Therefore, the product will have an inversion of stereochemistry as (<em>S</em>)-2-bromobutane is formed.

This product is treated with NaCN and CN⁻ is a very good nucleophile. The bromo-substitutent is a good leaving group, therefore, the nucleophile will do an sn2 substitution which inverts the stereochemistry once more to give the optically active product shown.

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3 years ago