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Natali [406]
3 years ago
10

What makes up the cell menbrane

Chemistry
2 answers:
Oxana [17]3 years ago
8 0

Explanation:

the principal components of plasma membrane are lipids phospholipid and cholesterol proteins and carbohydrates

Vikentia [17]3 years ago
5 0

With few exceptions, cellular membranes — including plasma membranes and internal membranes — are made of glycerophospholipids, molecules composed of glycerol, a phosphate group, and two fatty acid chains. Glycerol is a three-carbon molecule that functions as the backbone of these membrane lipids.

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3. What are vaccines? Are there different types?​
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Answer:Four types of vaccines are currently available: Live virus vaccines use the weakened (attenuated) form of the virus. The measles, mumps, and rubella (MMR) vaccine and the varicella (chickenpox) vaccine are examples

4 0
3 years ago
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Formula compound that forms between sodium and chlorite ions?
VashaNatasha [74]

Answer:

NaCl

Explanation:

Thus, for the compound between Na + and Cl −, we have the ionic formula NaCl (Figure 3.5 "NaCl = Table Salt").

6 0
2 years ago
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A scientist needs to collect a 0.050 mole sample of helium, but needs to know how large his helium container should be. What vol
Mumz [18]

We can use the ideal gas equation:

PV = nRT

P = 202.6kPa = 202600 Pa (You have to multiply by 1000)

n = 0.050 mole

R = 0.082 atm*l/(K*mol)

T = 400K

We will have to convert from Pa to atm or viceversa.

101325 Pa________1 atm

202600 Pa________x = 2.00 atm

2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K

V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm = 0.82 liters = 820 mililiters



8 0
4 years ago
A representative particle of carbon dioxide is
KonstantinChe [14]
Carbon atom with iron and helium
6 0
3 years ago
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g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
3 years ago
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