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2 years ago

The aqueous fluid outside of a cell is known as

Chemistry

2 answers:

Lera25 [3.4K]2 years ago

5 0

Answer:

<h2><u>ᎪꪀsωꫀᏒ </u></h2>

O interstitial

Explanation:

hope it's helpful to you

GrogVix [38]2 years ago

5 0

The answer is: Interstitial. Intracellular is wrong - it refers to inside of the cell.

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You are carefully watching the temperature of your melting point apparatus as it is heating up. At 132 C it is still a white sol

Lisa [10]

Answer:

See the answer below

Explanation:

<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>

1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.

2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.

6 0

3 years ago

Which type of erosion most likely formed the Grand Canyon?

butalik [34]

Flowing water of course

4 0

3 years ago

Ethylene gas and steam at 320°c and atmospheric pressure are fed to a reaction process as an equimolar mixture. the process prod

Reil [10]

The heat transfer formula is;
Q = m * c * Δ T >>>> (1)
where, Q is the heat transfer
m = mass (gram)
c = the specific heat capacity (J/g)
Δ T = change in temperature
∵ we have one mole of Ethanol
∴ the weight of ethanol equals its molecular weight = (2*12)+(6*1)+(16) = 46 g
we will assume that the specific heat capacity of ethanol is 2.46 J/g (from google)
ΔT = 25 - 320 = - 295 C
By substitution in (1)
∴ Q = 2.46 * 46 * (-295) = - 33382.2 J

4 0

3 years ago

Calculate the density (in g/mL) of a liquid that has a mass of 0.115g and a volume of 0.000265 L

notsponge [240]

density is mass/volume so we do

.115/.000265

433.962 g/mL

5 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago