<span>1) Use the balanced chemical equation to find the molar ratios (proportions) of each product and reactant.

3N2H4(l)→4NH3(g)+N2(g)

=> molar ratios: 3 mol N2H4 : 4 mol NH3

2) Use the product to reactant molar ratio, and the quantity of reactant to determine the yield:

2.0 mol N2H4 * [4mol NH3] / [3mol N2H4] = 2*4/3 mol NH3 = 2.7 mol NH3

Answer: 2.7 mol

</span>

8 0

3 years ago

The mass percent of carbon in a typical human is 18%, and the mass percent of 14C in natural carbon is 1.6 × 10-10%. Assuming a

Tema [17]

Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

$\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g$

Carbon-14 is $1.6\times 10^{-10}\ \%$ of the carbon in the body. So,

$\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g$

Also,

14 g of Carbon-14 contains $6.023\times 10^{23}$ atoms of carbon-14

So,

$2.48\times 10^{-8}\ g$ of Carbon-14 contains $\frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8}$ atoms of carbon-14

Atoms of carbon-14 = $1.07\times 10^{15}$

Given that:

Half life = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.00012 years⁻¹

Also, 1 year = $3.154\times 10^7$ s

So, The rate constant, k = $\frac{0.00012}{3.154\times 10^7}$ s⁻¹ = $3.8\times 10^{-12}\ s^{-1}$

Thus, decay events per second = $K\times atoms decayed$ = $3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s$ = 4066 decay/s

4 0

3 years ago

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

2 years ago