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2 years ago

Assuming it behaves as an ideal gas, calculate the density of sulfur dioxide, so2, at stp.

2 answers:

6 0

Hello!

At Standard Pressure and Temperature, an ideal gas has a molar density of 0,04464 mol/L.

So, we need to apply a simple conversion factor to calculate the density of Sulfur Dioxide using the molar mass of Sulfur Dioxide.

$\frac{0,04464 mol SO_2}{1 L SO_2}* \frac{64,066 g SO_2}{1 mol SO_2}=2,8599 g/L$

So, the Density of Sulfur Dioxide (SO₂) at STP is 2,8599 g/L

Have a nice day!

kolbaska11 [484]2 years ago

4 0

Answer:

2.86 g/L

Explanation:

Density can be defined as the ratio of mass to volume, that is, Mass/volume.

The molar mass of sulfur dioxide, SO2 can be calculated and it is given as: 32.07 + (16×2) = 64.07 grams per mole.

At standard temperature and pressure(step), one mole of an ideal gas is equal to the volume 22.4 litres. Hence, the gas density can be calculated using the formula below;

Density of sulfur dioxide,SO2 at stp= [molar mass(in grams per mole)÷ 1 mole] × [ 1 mol÷ 22.4 litres] ----------------------------------------------(***).

substitute in the values into the equation above, we have;

==> (64.07 grams per mol/ 1 mol) × 1 mol/22.4 litres.

===> 2.8603 grams per litre.

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For a lake in Michigan, researchers have determined that largemouth bass feed on smaller fish, which in turn feed on zooplankton

allsm [11]

Answer:

the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

Explanation:

Given the data in the question;

In 2016, there was 600,000 kg of zooplankton in the lake.

In 2017, an accidental runoff of insecticide near the lake caused a 50 percent decline of the zooplankton population in the lake.

Now,

The remaining mass of zooplankton after the 50% decline will be;

⇒ 600,000 kg zooplankton × 50%

⇒ 600,000 × 50/100

⇒ 300,000 kg of zooplankton

Now, with 10 percent trophic efficiency; smaller fish directly feed on zooplankton; decline in smaller fish mass will be;

⇒ 300,000 kg × 10%

⇒ 300,000 × 10/100

⇒ 30000 kg

Finally, with 10 percent trophic efficiency, largemouth bash directly feed on smaller fish, so the expected decline in mass of largemouth bash will be;

⇒ 30000 kg × 10%

⇒ 30000 kg × 10/100

⇒ 3,000 kg

Therefore, the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

7 0

2 years ago

Hospital patients are administered oxygen from an pressurized

Nataly_w [17]

Answer:

110L

Explanation:

Boyle's Law states that P1×V1=P2×V2

Volume is indirectly proportional to Pressure so P×V is constant

P1=55atm

V1=6L

P2=3atm

V2 is to be found

P1×V1=P2×V2

6×55=3×V2

330=3×V2

Answer: V2=110L

8 0

2 years ago

A graduated cylinder was filled with 25.0 mL of water. A solid object was immersed in the cylinder, raising the water level to 3

yawa3891 [41]

Answer:

Density of the object is 8759.494 grams/L

Explanation:

As we know density of an object is mass of the object divided by its volume

Given

Volume of the object is equal to the change in volume of water with in the cylinder when the object was immersed in water.

Hence, volume of object is equal to

$32.9 -25 = 7.9$ mL

Mass of the object is $69.2$ grams

Density

$= \frac{69.2}{7.9 *10^{-3}}\\= 8759.494$ grams/L

Density of the object is 8759.494 grams/L

7 0

2 years ago

Is zinc an example of a element or compound?

Lerok [7]

Answer:

Zinc is an <em>element</em>.

Explanation:

Compounds are composed of two or more elements

3 0

1 year ago

Read 2 more answers

what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?

MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻ + H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF, the [ OH ⁻] from the given pH, and finally the mass needed to produce that OH⁻ concentration.

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is 6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14 ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

F⁻ HF OH⁻

Equilibrium X 2.51 x 10⁻⁶ 2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X = 1.5 x 10⁻¹¹ ⇒ X = (2.51 x 10⁻⁶)² / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L * 0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF * 42.0 g/mol NaF = 6.03 g NaF

7 0

2 years ago