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In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

Arada [10]

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}$

$[A]=0.529M$

7 0

3 years ago

How many atoms of nitrogen are in the chemical formula Ni(NO3)2?

katen-ka-za [31]

Answer:

2 atoms of nitrogen are present.

4 0

2 years ago

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Select all the correct answers.

stich3 [128]

Answer : The products of the acid-base reaction between $HClO_3$ and LiOH are, $LiClO_3$ and $H_2O$

Explanation :

Neutralization reaction : It is a type of chemical reaction in which an acid react with a base to give salt and water as a product that means it reacts to give a neutral solution. It is also known as acid-base reaction.

The acid-base reaction will be:

$HClO_3+LiOH\rightarrow LiClO_3+H_2O$

Therefore, the products of the acid-base reaction between $HClO_3$ and LiOH are, $LiClO_3$ and $H_2O$

6 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

2 years ago

Fill in the coefficients that will balance the following reaction:

Sedaia [141]

the equation balanced is

3 PbI4 + 4 CrCl3 → 3 PbCl4 + 4 CrI3

8 0

2 years ago

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