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3 years ago

Increasing the intensity and duration of an exercise program too quickly can lead to injury.

2 answers:

6 0

It is true that a sudden increase in both intensity and duration of physical exercise can result in injury. The body needs time to process any changes. Muscles need to build strength and flexibility gradually and the joints need time to acclimate to changes in pressure.

iVinArrow [24]3 years ago

5 0

this is a true statement

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The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for

tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. $K(s)+O_2(g)\rightarrow KO_2(s)$

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$

Answer: A. $K(s)+O_2(g)\rightarrow KO_2(s)$ : decreases

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g)$ : decreases

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ : no change

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction $K(s)+O_2(g)\rightarrow KO_2(s)$ , as gaseous reactant is changed to solid product, entropy decreases.

In reaction $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0

3 years ago

What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?

Maru [420]

Answer:

There is 17,114825 g of powdered drink mix needed

Explanation:

Step 1 : Calculate moles

As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L

Since the volume is 100mL, we have to convert the concentration,

⇒0.5 / 1 = x /0.1 ⇒ 0.5* 0.1 = x = 0.05 M

This means there is 0.05 mol per 100mL

Step 2 : calculate mass of the powdered drink

here we use the formula n (mole) = m(mass) / M (Molar mass)

⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.

0.05 mol = mass / 342.2965 g/mol

To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g

There is 17,114825 g of powdered drink mix needed

3 0

3 years ago

Read 2 more answers

Pls help ill mark as brainliest :)

Alina [70]

Answer:

I think its a double reaction

Explanation:

4 0

3 years ago

___________refers to the amount of water vapor in the air.

ollegr [7]

Answer:

b humifity

Explanation:

4 0

3 years ago

Read 2 more answers

A gas sample occupies 3.25 liters at 297.5K and 2.4 atm. Determine the temperature at which the gas will occupy 4.25 L at 1.50 a

lorasvet [3.4K]

For equal moles of gas, temperature can be calculated from ideal gas equation as follows:

P×V=n×R×T ...... (1)

Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.

2.4 atm ×3.25 L=n×R×297.5 K

Rearranging,

n\times R=0.0262 atm L/K

Similarly at final pressure and volume from equation (1),

1.5 atm ×4.25 L=n×R×T

Putting the value of n×R in above equation,

1.5 atm ×4.25 L=0.0262 (atm L/K)×T

Thus, T=243.32 K

7 0

3 years ago