The normal boiling point of chloroform, which has a higher vapor pressure than water at 100 c, is ___

Which describes a chemical reaction? A. Raw eggs are fried in a pan B. Raw eggs are scrambled with a fork C. Raw eggs are placed

Otrada [13]

Raw eggs are scrambled with a fork

8 0

3 years ago

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I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

5 0

2 years ago

Which statement accurately describe the two animals?

ANEK [815]

The elephant has more inertia than the dog so it takes more force to change the motion of the elephant.

8 0

3 years ago

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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.

7 0

3 years ago

How many grams of tin are found in 3.5 moles of tin

Debora [2.8K]

Tin has a molecular weight of 118.7 That is the mass in grams of one mole.
<span>3.50 * 118.7 = 415</span>

7 0

2 years ago

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The normal boiling point of chloroform, which has a higher vapor pressure than water at 100 c, is ____.