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sergeinik [125]
3 years ago
15

In a phospholipid the phosphate head is? and the acid tail is

Chemistry
1 answer:
Marina CMI [18]3 years ago
6 0
The phosphate <u>head </u>is hydrophilic (faces outward in the lipid bilayer membrane), and the acid <u>tail </u>is lipophilic/hydrophobic (faces inward in the lipid bilayer membrane).
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A reaction that combines simpler reactants to form a new compound is called a
uranmaximum [27]

Answer:

A reaction that combines simpler reactants to form a new compound is called a

<h2>Synthesis reaction.</h2>
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3 years ago
Some hormones enter cells via
pochemuha
ANSWER: receptor-meditated endocytosis
I’m not sure if this is what you’re looking for but this is correct. Next time please add the full question and possible answers. HOPE this helps someone please leave thanks if so.
5 0
3 years ago
The mass of an electron is 9.11× 10–31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 106 m/s with
Kruka [31]
Just use the Heisenberg Uncertainty principle: 

<span>ΔpΔx = h/2*pi </span>

<span>Δp = the uncertainty in momentum </span>
<span>Δx = the uncertainty in position </span>
<span>h = 6.626e-34 J s (plank's constant) </span>

<span>Hint: </span>

<span>to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that </span>

<span>Δp = mv*(0.01) </span>

<span>m = mass of electron </span>
<span>v = velocity of electron </span>

<span>Solve for Δx </span>

<span>Δx = h/(2*pi*Δp) </span>

<span>And that is the uncertainty in position. </span>
6 0
3 years ago
1. Two boys balanced steady in a sea-saw game
anygoal [31]
Mutual

They are balanced steadily which means they’re at the same point
4 0
2 years ago
Read 2 more answers
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
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