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3 years ago

In Males which of the following organ performs a dual function of ejaculation of sperm and voiding of urine

Chemistry

2 answers:

Nat2105 [25]3 years ago

5 0

Answer: Urethra

Explanation:

The urethra is a tube like structure which performs the function of excretion as well as copulation. It carries the urine stored inside the bladder to the outside the body. During copulation, the semen is ejected from the urethra inside the female external genitalia that is vaginal canal. The urethra prevents the flow of urine and it just allows the ejaculation of semen during copulation.

Anna11 [10]3 years ago

3 0

I believe it is the urethra

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Name two example of oxygenase

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Air sksksks and I oop-

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3 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

3 years ago

When an electron falls from an excited state to ground state, is energy released or absorbed?

strojnjashka [21]

Answer:

Energy is released.

Explanation:

When an electron absorbs energy, it moves up into an excited state. When it releases energy, it will return to the ground state.

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2 years ago

Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a

nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

= 75.3%

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3 years ago

Details about outer ear

Verdich [7]

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