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3 years ago

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Which of theCardiorespiratory fitness is most likely to improve due to which of the following types of exercise? A. bicycling B.

weight lifting C. push-ups D. sit-ups Please select the best answer from the choices provided. A B C D

1 answer:

gizmo_the_mogwai [7]3 years ago

7 0

Answer: A (Bicycling)

Explanation:

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When the reaction takes place,2 grams of carbon dioxide will be made but vaporized into air.

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An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
<span>delta U is change in internal energy </span>

<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>

<span>delta U = q + w </span>

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<span>q = m x c x delta T </span>

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</span>
Hope this Helps! Have A Wonderful Day! :)

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A cell that reproduces

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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

<u>Calculating for first dilution:</u>

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

<u>Calculating for second dilution:</u>

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

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3 years ago