All categories

3 years ago

What are the effects of viagra?

2 answers:

7 0

An erection that will not go away (priapism).
Sudden vision loss in one or both eyes
Sudden hearing decrease or hearing loss.

mestny [16]3 years ago

5 0

Viagra is used to stifle the messengers of cGMP. So that a male can sustain an erection.

Side effects can include headaches, stuffy nose, impaired vision, sensitivity to light, indigestion, and the possibility of a (painful)long lasting erection

You might be interested in

Which item is made from an alloy? A. steel tray B. glass plate C. credit card D. copper wire

Naya [18.7K]

Not sure if this is correct but i would choose the steel tray.

6 0

3 years ago

Read 2 more answers

what would you be most likely to measure by immsering and object in water and seeing how much the water level rises

Katena32 [7]

Mass, if you know what element you are working with.

4 0

3 years ago

Which statement best characterizes chemical reactions

amid [387]

Dangerous
hope i helpedXD

5 1

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

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

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

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

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

3 years ago

At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249

oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= $\frac{1249 g}{396.35 g/mol}$

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

7 0

3 years ago