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3 years ago

11

IF 500 mL of air contains 2.50x10^22 particles (atoms and molecules). how much particles do you inhale in one day if you breath

15000L of air?

1 answer:

vaieri [72.5K]3 years ago

8 0

Answer:

I think it is 0.00192

Explanation:

28.795705

15000

=0.00192

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This biome has the nutrient poor soil

nikklg [1K]

Answer:

the answer is the tropical rain forest

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3 years ago

g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u

DIA [1.3K]

Answer:

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

m → molality (moles of solute in 1kg of solvent)

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg . 0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

4 0

4 years ago

The density of liquid mercury is 13.69 g/cm^3. How many atoms of mercury are in a 15.0 cm^3 sample? Use "E" for "x10" and use si

balu736 [363]

The formula is m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
1 mole = 6.022 E23 atoms
0.97 mole=?

6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms

6 0

3 years ago

Name the reaction type & mechanism of the following reactions:

Natali5045456 [20]

Answer:

MECHANISM:

1) The lone pair on oxygen attacks the H-Br molecule forming a hydronium ion.

2) Formation of carbocation.

3) Attack of Nucleophile Br − .

Explanation:

4 0

3 years ago

Read 2 more answers

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

3 years ago