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AfilCa [17]
3 years ago
9

How do carrier proteins transport substances across cell membranes?

Chemistry
1 answer:
stepan [7]3 years ago
7 0
<em>Transport of substances across cell membrane:</em>
<em>Carrier proteins facilitate transport of material across a membrane by changing its shape to allow a substance to pass through its membrane. It transports substances from high concentration to lower concentration levels.</em>
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Which is a way to Express concentration of solution.?​
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Explanation:

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write the equilibrium expression for the base ionization of the weak base of methylamine, ch3 nh2 . hint: first write the balanc
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CH3NH2 + HOH ==> CH3NH3^+ + OH^-Which molecule/ion accepts a proton. That is the base. Which molecule/ion donates a proton. That is the acid.

A stable subatomic particle known by the symbol for "proton"

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1 year ago
How many electrons are there in an atom of carbon?​
Mashutka [201]

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if its a neutral atom there should be 6 electrons because there are 6 protons

Explanation:

8 0
3 years ago
Does Hornfels have biological origins
anzhelika [568]

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3 0
3 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
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