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kow [346]
3 years ago
8

Nayla grew skin in a lab by adding cells to a synthetic material. The skin functioned normally for 12 days. Then Nayla separated

some
of the skin cells into cell membranes, cytoplasm, and vacuoles to studythem. WhichofthefollowingwerealiveduringNayla’s experiment?
a. The skin and the cytoplasm
b. The skin and the skin cells
c. The cell membranes and the skin cells
d. The cell membranes and the cytoplasm
Chemistry
1 answer:
iris [78.8K]3 years ago
6 0

Answer:B) The Skin and The Skin cells

Explanation:

Hope this helped

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What is the PH of 8.1x10^-5 M HCLO4 solution
Ivanshal [37]
HCl04 is a strong acid solution. So, in theory everything dissociates.

pH= -log(0.000081)= 4.09
6 0
3 years ago
PLEASE HELPPPP! HOW DO I DO THIS???
kow [346]

Answer:

1552.83J Released

Explanation:

1. mass/m=225

Initial temp:86C, final:32.5C

Changed Temp: 32.5-86= -53.5C

s=0.129 J/gC

Formula: q= m times s times changed Temp.

q=(225)(0.129)(-53.5)

q= -1552.83 J

q=1552.83 J Released

8 0
3 years ago
What is the partial pressure (in mmHg) of oxygen in a sample of sea level air (a mixture
yaroslaw [1]

The partial pressure of oxygen, P(O₂) is 198.83 mmHg.

<h3>What is partial pressure of a gas?</h3>

The partial pressure of a gas is the pressure gas will exert in a mixture of gases which do not react chemically together.

The sum of partial pressure of gases at atmospheric pressure = 760 mmHg

P(N₂) +  P(H₂O) + P(CO₂) + P(O₂) + P(other) = 760 mmHg

P(N₂) = 0.72 atm = 547.2 mmHg

P(H₂O) = 7.7 torr = 7.7 mmHg

P(CO₂) = 0.37 mmHg

P(other) = 0.97 kPa = 5.9 mmHg

P(O₂) = 760 - (547.2 + 7.7 + 0.37 + 5.9) = 198.83 mmHg

Therefore, the partial pressure of oxygen, P(O₂) is 198.83 mmHg.

Learn more partial pressure at: brainly.com/question/14119417

#SPJ1

4 0
2 years ago
10. A 20.00 mL sample of 0.150 mol/L ammonia (NH3(aq)) is titrated to the equivalence point by 20.0 mL of a solution of 0.150 mo
Natalija [7]

Answer:

\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}

Explanation:

a) Balanced equation

The balanced chemical equation for the titration is

\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq)}

b) pH at start

For simplicity, let's use B as the symbol for NH₃.

The equation for the equilibrium is

\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}

(i) Calculate [OH]⁻

We can use an ICE table to do the calculation.

                      B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹:     0.150               0         0

C/mol·L⁻¹:       -x                 +x       +x

E/mol·L⁻¹:  0.150 - x            x          x

K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}

Check for negligibility:

\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150

(ii) Solve for x

\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}

(iii) Calculate the pH

\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}

(c) pH at equivalence point

(i) Calculate the moles of each species

\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}

                 B    +    HI   ⇌   BH⁺ + I⁻

I/mol:       3.00    3.00         0

C/mol:    -3.00   -3.00     +3.00

E/mol/:       0          0          3.00

(ii) Calculate the concentration of BH⁺

At the equivalence point we have a solution containing 3.00 mmol of NH₄I

Volume = 20.00 mL + 20.00 mL = 40.00 mL

\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}

(iii) Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

                      BH⁺+ H₂O ⇌ H₃O⁺ +  B

I/mol·L⁻¹:     0.0750                 0        0

C/mol·L⁻¹:        -x                     +x      +x

E/mol·L⁻¹:   0.0750 - x             x         x

K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ $\ll$ 0.0750}

\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}

(iv) Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}

The titration curve below shows the pH at the beginning and at the equivalence point of the titration.

8 0
3 years ago
1s² 2s² 2p⁶ que elemento es? ​
lana66690 [7]

Answer:

Esta configuración electrónica pertenece al elemento, neón.

Si tienes preguntas, házmelo saber . :)

5 0
3 years ago
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