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3 years ago

Where does fertilization occurs in human beings

Chemistry

2 answers:

Harman [31]3 years ago

7 0

In human beings fertilization occurs in fallopian tube

vladimir1956 [14]3 years ago

5 0

Answer:

Fertilization usually takes place in a fallopian tube that links an ovary to the uterus. If the fertilized egg successfully travels down the fallopian tube and implants in the uterus, an embryo starts growing.

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We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of

Arte-miy333 [17]

Answer:

A. pH using molar concentrations = 2.56

B. pH using activities = 2.46

C. pH of mixture = 2.56

Explanation:

A. pH using molar concentrations

ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺

HA + H₂O ⇌ A⁻ + H₃O⁺

We have a solution of 0.08 mol HA and 0.04 mol A⁻

We can use the Henderson-Hasselbalch equation to calculate the pH.

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}$

B. pH using activities

(i) Calculate [H⁺]

pH = -log[H⁺]

$\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}$

(ii) Calculate the ionic strength of the solution

We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.

The formula for ionic strength is

$I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041$

(iii) Calculate the activity coefficients

$\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79$

(iv) Calculate the initial activity of A⁻

a = γc = 0.79 × 0.04= 0.032

(v) Calculate the pH

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\$

C. Calculate the pH of the mixture

The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.

The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.

The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.

The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.

(i) Calculate the ionic strength

$I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10$

(ii) Calculate the activity coefficients

$\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69$

(iii) Calculate the initial activity of A⁻:

a = γc = 0.69 × 0.05= 0.034

(iv) Calculate the pH

$\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}$

3 0

3 years ago

a gas thermometer measure temperature by measuring the pressure of a gas inside the fixed volume container a thermometer reads a

sdas [7]

Hello!

a gas thermometer measure temperature by measuring the pressure of a gas inside the fixed volume container a thermometer reads a pressure of 248 torr at 0 degrees Celsius what is the temperature when the thermometer reads a pressure of 345 torr

We have the following information:

P1 (initial pressure) = 248 torr

T1 (initial temperature) = 0 ºC (in Kelvin)

TK = TºC + 273.15 → TK = 0 + 273.15 → T1 (initial temperature) = 273.15 K

P2 (final pressure) = 345 torr

T2 (final temperature) = ? (in Kelvin)

According to the Law of Charles and Gay-Lussac in the study of gases, we have an isochoric (or isovolumetric) transformation when its volume remains constant or equal, then we will have the following formula:

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

$\dfrac{248}{273.15} = \dfrac{345}{T_2}$

$248*T_2 = 345*273.15$

$248\:T_2 = 94236.75$

$T_2 = \dfrac{94236.75}{248}$

$T_2 = 379.9868952... \to \boxed{\boxed{T_2 \approx 380\:K}}\end{array}}\qquad\checkmark$

If you want the solution in Celsius Temperature, we have:

TC = TK - 273.15

TC = 380 - 273.15

TC = 106.85 → Temperature = 106.85 ºC

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

6 0

3 years ago

How much energy in joules is released when 18.5 grams of copper cools from 285 °C down to 45 °C

nadya68 [22]

Answer: 1709.4 Joules

Explanation:

The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = ?

M = 18.5 grams

Recall that the specific heat capacity of copper C = 0.385 J/g.C

Φ = 285°C - 45°C = 240°C

Then, Q = MCΦ

Q = 18.5grams x 0.385 J/g.C x 240°C

Q = 1709.4 Joules

Thus, 1709.4 Joules is released when copper is cooled.

5 0

3 years ago

Consider the following intermediate chemical equations.

vfiekz [6]

Answer:

-250.3kJ

Explanation:

Based in the reactions and using -Hess's law-:

(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ

(2) 4PCl₅(g) → P₄(s) + 10Cl₂ ΔH₂ = 3438kJ

The sum of (1) + (2) is:

4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ

Dividing this reaction in 4:

PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = -250.3kJ

8 0

3 years ago

The atom fluorine generally will become _____ stable through the formation of an ionic chemical compound by accepting _______ e

larisa86 [58]

The atomic number of Fluorine is 9

Valence (outer) electron configuration is : 2s²2p⁵

Therefore, it requires 1 electron in the p-orbital to complete its octet of 8 electrons.

Thus, the atom Fluorine generally will become more stable through the formation of an ionic chemical compound by accepting 1 electron from another atom. This process will fill its outer energy level.

Ans: A) more, 1

3 0

3 years ago

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Where does fertilization occurs in human beings​

Where does fertilization occurs in human beings