Calculate the<br> charge of 100 electrons (in coulombs)

Which of the following is not a component of fitness? A. Cardiovascular fitness B. Muscular endurance C. Pulmonary endurance D.

Ne4ueva [31]

Answer:

C. Pulmonary endurance

Explanation:

I'm pretty sure it's "C" because cardiovascular and pulmonary endurance are the same thing and usually you'd hear cardiovascular more than pulmonary.

Sorry if I'm wrong!

7 0

3 years ago

During the first stage of sleep deprivation, the subject may __________.

klio [65]

<h2>First stage of sleep Deprivation Subject </h2>

During the first stage of sleep deprivation, the subject is NREM which stands for non-rapid eye movement. In this condition, we are not sleeping in the depth. It can be said as dreamless sleep. On electro-encephalography recording, the brain waves are not fast so they have high voltage.

In this condition, the breathing, heart rate and blood pressure is low. The sleep is comparatively tranquil. NREM lasts for 90 minutes to 120 minutes. It accounts for about 75% of the normal sleep time. Rapid eye movements do not occur.

7 0

3 years ago

Read 2 more answers

Place the following structures in the appropriate category based on their location in a cell

Studentka2010 [4]

Answer:

Cell surface.

Cilium
Plasma membrane
Microvilus.

Nucleus

Cellular DNA
Nucleolus
Chromosome.

Cytoplasm

Mitochondria
Golgi apparatus
Cytoskeleton.

3 0

3 years ago

Which statement is true about the formation of bonds?

Ivenika [448]

There are no true statements on the list you attached.

7 0

3 years ago

Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of

MariettaO [177]

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of $Cl_2$ = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of $Cl_2$ = 71 g/mole

Molar mass of $PCl_5$ = 208.24 g/mole

First we have to calculate the moles of $P$ and $Cl_2$ .

$\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles$

$\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2P+5Cl_2\rightarrow 2PCl_5$

From the balanced reaction we conclude that

As, 5 moles of $Cl_2$ react with 2 moles of $P$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $P$

That means, in the given balanced reaction, $Cl_2$ is a limiting reagent and it limits the formation of products and $P$ is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of $PCl_5$ .

As, 5 moles of $Cl_2$ react with 2 moles of $PCl_5$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $PCl_5$

Now we have to calculate the mass of $PCl_5$ .

$\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5$