Please help me with this science question

As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.71 atm and a temperature of 25.0 C,

Semenov [28]

Answer:

$1.32\times 10^{-3} mol/L$ is the solubility of nitrogen gas in a diver's blood.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

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

$C_{N_2}=K_H\times p_{liquid}$

where,

$K_H$ = Henry's constant = $6.26 \times 10^{-4}mol/L.atm$

$p_{N_2}$ = partial pressure of nitrogen

$p_{N_2}= P\times \chi_{N_2}$ (Raoult's law)

$=2.71 atm\times 0.78=2.1138 atm$

$C_{N_2}=6.26 \times 10^{-4}mol/L.atm\times 2.1138 atm$

$C_{N_2}=1.32\times 10^{-3} mol/L$

$1.32\times 10^{-3} mol/L$ is the solubility of nitrogen gas in a diver's blood.

3 0

4 years ago

Water flows over Niagara Falls at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing

Zinaida [17]

Answer:

1) The power of Niagara Falls is 1.176 × 10⁹ W

2) The number of 15 W LED light bulbs it could power is 78.4 × 10⁶ light bulbs

Explanation:

1) The Niagara falls water mass flow rate = 2,400,000 kg/s

The height of the fall = 50 meters

The gravitational potential energy = Mass (kg) × height (m) × gravity (9.8 m/s²)

The power = The energy converted per second = Mass flow rate (kg/s) × height (m) × gravity (9.8 m/s²)

Therefore;

The power of Niagara Falls= 2,400,000 kg/s × 50 m ×9.8 m/s²= 1.176 × 10⁹ W

The power of Niagara Falls = 1.176 × 10⁹ W

2) The number, n, of 15 W LED light bulbs it could power is given by the relation;

n × 15 W = 1.176 × 10⁹ W

∴ n = 1.176 × 10⁹ W/(15 W) = 78.4 × 10⁶ light bulbs

The number of 15 W LED light bulbs it could power = 78.4 × 10⁶ light bulbs.

6 0

3 years ago

What are some examples of supersaturated solutions? Give 10.

Phantasy [73]

Examples include carbonated water (i.e. soda water); honey; sugar syrup (used in confectionery); supersaturated drug delivery systems. "SDDS"; and sodium acetate solutions prepared from 160 g NaOAc and 30 mL water.

7 0

3 years ago

A reaction between liquid reactants takes place at −10.0°C in a sealed, evacuated vessel with a measured volume of 45.0L. Measur

maks197457 [2]

Answer: 0.0624 atm

Explanation:-

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = ?

V= Volume of the gas = 45.0 L

T= Temperature of the gas = -10.0°C = 263 K $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

Moles of gas= $\frac{\text{ given mass}}{\text{ molar mass}}= \frac{19.0g}{146g/mole}=0.130moles$

$P=\frac{nRT}{V}=\frac{0.130\times 0.0821\times 263}{45.0}=0.0624atm$

The pressure of sulfur hexafluoride gas in the reaction vessel after the reaction is 0.0624 atm

3 0

3 years ago

In the reaction of 675.9 grams of barium chloride and excess silver(I) nitrate, how many grams of silver(I) chloride should get

zvonat [6]

Taking into account the reaction stoichiometry, the mass of silver(I) chloride formed is 930.37 grams.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + BaCl₂ → 2 AgCl + Ba(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AgNO₃: 2 moles
BaCl₂: 1 mole
AgCl: 2 moles
Ba(NO₃)₂: 1 mole

The molar mass of the compounds is:

AgNO₃: 169.87 g/mole
BaCl₂: 208.24 g/mole
AgCl: 143.32 g/mole
Ba(NO₃)₂: 261.34 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

AgNO₃: 2 moles ×169.87 g/mole= 339.74 grams

BaCl₂: 1 mole ×208.24 g/mole= 208.24 grams

AgCl: 2 moles ×143.32 g/mole= 286.64 grams

Ba(NO₃)₂: 1 mole×261.34 g/mole= 261.34 grams

<h3>Mass of silver(I) chloride formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 208.24 grams of barium chloride form 286.64 grams of silver(I) chloride, 675.9 grams of barium chloride form how much mass of silver(I) chloride?

$mass of silver(I) chloride=\frac{675.9 grams of barium chloride x286.64 grams of silver(I) chloride }{208.24 grams of barium chloride}$

mass of silver(I) chloride= 930.37 grams

Finally, the mass of silver(I) chloride formed is 930.37 grams.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

8 0

2 years ago