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3 years ago

How many total atoms are in 0.560 g of P2O5

Chemistry

1 answer:

ioda3 years ago

7 0

Answer:

16.499 × 10∧ 21 atoms

Explanation:

Given data:

mass of P2O5= 0.560 g

number of atoms= ?

first of all we will calculate the molar mass of P2O5:

P = 2×31 g/mol = 62 g/mol

O = 5× 16 = 80 g/mol

molar mass of P2O5 = 142 g/mol

Noe we will find the moles of 0.560 g P2O5:

moles = mass / molar mass

moles = 0.560 g/ 142 g/mol

moles = 0.0039 mol

now we will find the atoms present in 0.0039 moles:

0.0039 × 6.02 × 10∧ 23 molecules

2.357 × 10∧ 21 molecules

P2O5 consist of 7 atoms:

2.357 × 10∧ 21 × 7 = 16.499 × 10∧ 21 atoms

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Answer:

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3 years ago

if the density of a certain spherical atomic nucleus is 1.0x10^14 g cm^-3 and its mass is 2.0x10^-23 g, what is it radius in cm?

bagirrra123 [75]

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3 years ago

Read 2 more answers

Reynolds number, for an aorta of 0.9-centimeter diameter, calculate the blood flow in liters per minute when the flow regime in

olga55 [171]

Answer:

Flow in liters per minute is 4 L/min

Explanation:

For this case, we have an aorta of 0.9 cm of diameter (D). Let's suppose an uniform and constant diameter for calculation purposes.

D = (0.9 cm)(1m/100cm)

D = 0.009 m

It is required to calculate the blood flow in liters per minute when the flow regime changes from laminar to turbulent. Laminar flow is usually less than 2500 for Reynolds value, and Turbulent flow when Re is higher than 2500. So, we need to study the phenomenon for

Re = 2500.

Using the definition of Reynolds we can find the velocity average of the blood, and use it to find flow. Where blood density is $\rho$ , aorta diameter is D, average velocity is v and blood viscosity is $\mu$

$Re = \frac{\rho v D}{\mu } \\v = \frac{Re*\mu}{\rho D}$

From data problem, we have Re, D values. As we need blood density and blood viscosity we can find them in medical studies. For example: in this online document: Blood flow analysis of the aortic arch using computational fluid dynamics †

Satoshi Numata, Keiichi Itatani, Keiichi Kanda, Kiyoshi Doi, Sachiko Yamazaki, Kazuki Morimoto, Kaichiro Manabe, Koki Ikemoto, Hitoshi Yaku Author Notes

European Journal of Cardio-Thoracic Surgery, Volume 49, Issue 6, June 2016, Pages 1578–1585, https://doi.org/10.1093/ejcts/ezv459

Published: 20 January 2016

$\rho = 1060 kg/m^3\\\mu = 0.0004 kg/(ms)\\$

$v = \frac{Re*\mu}{\rho D}\\v = \frac{2500 * 0.004 kg/(ms)}{(1060 kg/m3)(0.009 m)}\\v = 1.04 m/s\\$

Average blood velocity is 1.04 m/s. After that, we can calculate the flow (Q) using the flow are (Ao) of aorta.

Q = v*Ao

$Q = (1.4m/s)*(\pi*(0.009m/2)^2)\\Q = 6.67 * 10^-5 \frac{m^3}{s}\\Q = (6.67 * 10^-5 \frac{m^3}{s})(\frac{1000 L}{1 m^3} ) (\frac{60 s}{1 min} )\\Q = 4 L/min$

Finally, the blood flow in liters per minute is 4 L/min when the flow regime in the aorta changes from laminar to turbulent.

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3 years ago

A person tries to pour exactly one kilogram of sand onto a scale on five

blsea [12.9K]

Answer:

A. While inconsistent, his average of 1.0 kg made his results quite

accurate

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sergiy2304 [10]

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Disagree

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3 years ago