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3 years ago

What organ is a muscle and expands and contracts to allow lungs to take in air

Chemistry

2 answers:

jeka57 [31]3 years ago

4 0

Answer:

Your diaphragm contracts and expands

Katyanochek1 [597]3 years ago

4 0

Answer:

The diaphragm

Explanation:

The diaphragm, located below the lungs, is the major muscle of respiration.

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What is the unit of Avogadro's number?

kompoz [17]

Answer:

it may be electron, atom or ions depend on the nature of the substance and the character of reaction

like :- particles per mole, coulombs per electron

8 0

3 years ago

40 cm3 of acid were mixed with 60 cm3 of alkali in an insulated container. What name is given to this type of reaction?

Stells [14]

Answer:

neutralization reaction

6 0

1 year ago

What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH)?

Vanyuwa [196]

Answer:

Mole fraction for solute = 0.1, or 10%

Molality = 6.24 mol/kg

Explanation:

22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH

Mass of solution = 100 g

Mass of solute = 22.3 g

Mass of solvent = 100 g - 22.3g = 77.7 g

Let's convert the mass to moles

22.3 g . 1mol/ 46 g = 0.485 moles

77.7 g. 1mol / 18 g = 4.32 moles

Total moles = 4.32 moles + 0.485 moles = 4.805 moles

Xm for solute = 0.485 / 4.805 = 0.100 → 10%

Molality → mol/ kg → we convert the mass of solvent to kg

77.7 g. 1 kg / 1000g = 0.0777 kg

0.485 mol / 0.0777 kg = 6.24 m

6 0

3 years ago

Read 2 more answers

Which describes the dissolving process? (Choose all that Apply)

stich3 [128]

Answer:

A and D

Explanation:

I took the test, and choose both B and C and got it wrong

4 0

3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago