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Iteru [2.4K]
3 years ago
12

Investigators studying a particular cell response are curious about what type of cell-surface receptor might trigger it. They in

cubate cells with EDTA, a chemical known to sequester Ca2+. This treatment, they find, abolishes the response they are studying. Based on this information, which type of receptor is likely to be involved?
Chemistry
1 answer:
belka [17]3 years ago
3 0

Answer:

An ion channel, more specifically a calcium channel.

Explanation:

The electrical activity of the cells is regulated by ion channels. Calcium channels, also referred as the voltage-gated calcium channels constitute one group of a superfamily of ion channels. A change in voltage across the membrane or small molecules triggers calcium channels to open, allowing calcium to flow into the cell. Inside the cell, calcium acts as a second messenger, it binds to calcium sensitive proteins to induce different responses and support several functions such as muscle contraction, hormone and neurotransmitter secretion, gene regulation, activation of other ion channels, control of action potentials, cell survival, etc.

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A covalent bond is likely to be polar when ________. one of the atoms sharing electrons is more electronegative than the other a
nekit [7.7K]

Answer:

one of the atoms sharing electrons is more electronegative than the other atom

Explanation:

3 0
3 years ago
Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

no of moles of F^- = 1 \times 10^{-6}\ mol

Total volume = 0.02 L

[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\  \\  \[[Ca^{2+}}] = 0.0005 \ mol/L

[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}

[F^{-}] = 5 \times 10^{-5}  \ mol/L

Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}

Since Q<ksp, then there will no be any precipitation of CaF2

3 0
3 years ago
__C2H4+__O2--&gt;__CO2+__H2O
mihalych1998 [28]

Answer:

Here is your answer mate :D

3 0
2 years ago
HELP ME PLEASE ILL MARK U THE BRAINLIEST
ElenaW [278]

Answer:

pentene-alkanes-C5H10

octane-alkanes-C 8H 18

butene-alkene-c4h8

decane-alkane-c10h22

6 0
3 years ago
1.72 moles of NOCI were placed in a 2.50 L reaction chamber
elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI  remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

   2NOCI(g) = 2NO(g) + Cl2(g).

I    1.72

C   0.56           0.56         0.28

E   1.16              0.56         0.28

Molarity at equilibrium :

NOCl :

\tt \dfrac{1.16}{2.5}=0.464

NO :

\tt \dfrac{0.56}{2.5}=0.224

Cl2 :

\tt \dfrac{0.28}{2.5}=0.112

\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026

4 0
3 years ago
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