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Blizzard [7]
3 years ago
8

Lipids that are liquid at room temperature are known as ____.

Chemistry
1 answer:
zaharov [31]3 years ago
7 0
Lipids that are liquid at room temperature are known as oils.
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Describe one function of your body and how it maintains homeostasis
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Explanation:

Maintenance of homeostasis usually involves negative feedback loops. These loops act to oppose the stimulus, or cue, that triggers them. For example, if your body temperature is too high, a negative feedback loop will act to bring it back down towards the set point, or target value, of 98.6 ∘ F 98.6.

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3 years ago
Answers please for all
iVinArrow [24]
14.
A. Lost a proton

b.Base

15.
A. Gain

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3 years ago
Balance the following equation. <br> C + H2 + 02 → C2H60
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4C + 6H2 + O2 -> 2C2H6O
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Read 2 more answers
Consider the reaction. A(aq) = 2 B(aq) Kc = 6.90 x 10 6 at 500 K If a 3.00 M sample of A is heated to 500 K, what is the concent
kirza4 [7]

Answer:

0.004548 M is the concentration of B at equilibrium at 500 K.

Explanation:

                         A(aq) ⇆ 2 B(aq)

Initially               3.00 M

At equilibrium   3.00 -x    2x

Equilibrium constant of the reaction at 500 K =K_c=6.90\times 10^{-6}

Concentration of A at 500 K at equilibrium , [A] = (3.00 -x )M

Concentration of B at 500 K at equilibrium,[B]= 2x

An expression of equilibrium constant is given as:

K_c=\frac{[B]^2}{[A]}

6.90\times 10^{-6}=\frac{4x^2}{(3.00-x)}

On solving for x:

x = 0.002274 M

[B] = 2 x = 2 × 0.002274 M = 0.004548 M

[A] = (3-x) = 3 M - 0.002274 M =2.997726 M

0.004548 M is the concentration of B at equilibrium.

8 0
3 years ago
Which of the following buffer systems would you use if you wanted to prepare a solution having a pH of approximately 9.5?Which o
zalisa [80]

Answer:

b. 0,08M NH₄⁺ / 0,12M NH₃

Explanation:

<em>The buffers are:</em>

<em>a. 0,08M H₂PO₄⁻ / 0,12M HPO₄²⁻</em>

<em>b. 0,08M NH₄⁺ / 0,12M NH₃</em>

It is possible to find out the pH of a buffer using Henderson-Hasselbalch formula:

<em>pH = pka + log₁₀ [A⁻] / [HA] </em><em>(1)</em>

Where A⁻ is the conjugate base of the weak acid, HA.

a. For the system H₂PO₄⁻ / HPO₄²⁻ pka is <u><em>7,198</em></u>. Replacing in (1)

pH = 7,198 + log₁₀ [0,12] / [0,08]

<em>pH = 7,37</em>

That means this buffer system don't have a pH of approximately 9,5

b. For the system NH₄⁺ / NH₃ pka is <em><u>9,3</u></em>. Replacing in (1)

pH = 9,3 + log₁₀ [0,12] / [0,08]

<em>pH = 9,50</em>

That means this buffer system is the buffer you need to use.

I hope it helps!

7 0
4 years ago
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