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3 years ago

9

What is the study of organism at the early stages of devolopment

1 answer:

egoroff_w [7]3 years ago

6 0

Answer: Embryology

Explanation:

You might be interested in

A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using this information, calculate Ka for HA.

slavikrds [6]

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant ( ${ \alpha}$ ) is :

$\qquad \sf \dashrightarrow \: \alpha = \frac{dissociation \: \: percentage}{100}$

$\qquad \sf \dashrightarrow \: \alpha = \frac{9.4}{100} = 0.094$

Now, at initial stage :

$\textsf{ Conc of HA = 0.200 M}$

$\textsf{Conc of H+ = 0 M}$

$\textsf{Conc of A - = 0 M}$

At equilibrium :

$\textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M}$

$\textsf{Conc of H+ = 0.094(0.200) = 0.0188 M}$

$\textsf{Conc of A - = 0.094(0.200) = 0.0188 M}$

Now, we know :

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}}$

( big brackets represents concentration )

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}}$

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}}$

$\qquad \sf \dashrightarrow \: { K_a \approx 0.00195 }$

$\qquad \sf \dashrightarrow \: {K_a \approx 1.9 × {10}^{-3} }$

7 0

2 years ago

Given the molar absorptivity for a species X of 1600 M-1cm-1 at a wavelength of 270 nm, and 400 M-1cm-1 at a wavelength of 540 n

Andre45 [30]

Answer:

Explanation:

From the given information:

At wavelength = 270 nm

$\varepsilon x_1 = 1600 \ m^{-1} \ cm^{-1} \\ \\ \varepsilon y_1 = 200 \ m^{-1} \ cm^{-1}$

At 270 nm

Suppose x is said to be the solution for the concentration of x and y to be the solution for the concentration of y;

Then:

$\varepsilon x_1 \ l + \varepsilon y_1 \ l= 0.5 \\ \\ A = A_1 + A_2$

$1600 xl + 200 yl= 0.5$

Divide both sides by 200

$8xl + yl = \dfrac{0.5}{200}$

$8x + y = \dfrac{0.5}{200}l$

Use l = 1cm (i.e the standard length)

Then;

$8x + y = \dfrac{0.5}{200} ---- (1)$

<u>For 540 nm:</u>

$\varepsilon x_2 x \ l + \varepsilon y_2 y \ l= 0.5 \\ \\ 40 xl + 800 yl = 0.5$

$x + 20 y = \dfrac{0.5}{400 \ l}$

since l = 1

$x + 20 y = \dfrac{0.5}{400 \ } --- (2)$

Equating both (1) and (2) together, we have:

$8x + y - 8x - 160 y = \dfrac{0.5}{200} - \dfrac{0.5 \times 8}{400} \\ \\ \implies - 159 y = \dfrac{0.5}{200} ( 1 - \dfrac{8}{2}) \\ \\ -159 y = \dfrac{-0.5 \times 3}{200} \\ \\ 159 \ y = 0.0075 \\ \\ y = \dfrac{0.0075}{159} \\ \\ y = 0.00004716 \\ \\ y = 4.7 \times 10^{-5 } \ M$

3 0

3 years ago

Kryger [21]

The empirical formula : C₁₂H₄F₇

The molecular formula : C₂₄H₈F₁₄

<h3>Further explanation</h3>

mol C (MW=12 g/mol)

$\tt \dfrac{18.24}{12}=1.52$

mol H(MW=1 g/mol) :

$\tt \dfrac{0.51}{1}=0.51$

mol F(MW=19 g/mol)

$\tt \dfrac{16.91}{19}=0.89$

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7

Empirical formula : C₁₂H₄F₇

(Empirical formula)n=molecular formula

( C₁₂H₄F₇)n=562 g/mol

(12.12+4.1+7.19)n=562

(281)n=562⇒ n =2

Molecular formula : C₂₄H₈F₁₄

6 0

3 years ago

phosphorus trifluoride is formed from its elements P4 (s) F2 (g) ---> PF3 (g) how many grams of fluorine are needed to react

Oksi-84 [34.3K]

This is an incomplete question, here is a complete question.

Phosphorus trifluoride is formed from its elements.

$P_4(s)+6F_2(g)\rightarrow 4PF_3(g)$

How many grams of fluorine are needed to react with 6.20 g of phosphorus?

Answer : The mass of $F_2$ needed are, 11.4 grams.

Explanation : Given,

Mass of $P_4$ = 6.20 g

Molar mass of $P_4$ = 124 g/mol

Molar mass of $F_2$ = 38 g/mol

First we have to calculate the moles of $P_4$

$\text{Moles of }P_4=\frac{\text{Given mass }P_4}{\text{Molar mass }P_4}$

$\text{Moles of }P_4=\frac{6.20g}{124g/mol}=0.05mol$

Now we have to calculate the moles of $F_2$

The balanced chemical equation is:

$P_4(s)+6F_2(g)\rightarrow 4PF_3(g)$

From the balanced reaction we conclude that

As, 1 mole of $P_4$ react with 6 moles of $F_2$

So, 0.05 moles of $P_4$ react with $0.05\times 6=0.30$ moles of $F_2$

Now we have to calculate the mass of $F_2$

$\text{ Mass of }F_2=\text{ Moles of }F_2\times \text{ Molar mass of }F_2$

$\text{ Mass of }F_2=(0.30moles)\times (38g/mole)=11.4g$

Therefore, the mass of $F_2$ needed are, 11.4 grams.

4 0

3 years ago

Explain how the re-introduction of wolves impacted the flow in yellowstone

zzz [600]

Answer:

more predators means less prey, it also probably affected other predators by introducing more territorial animals making it more crowded and harder to find a home for everyone

8 0

4 years ago