All categories

3 years ago

Describe how phospholipids form a barrier between water inside the cell and water outside the cell.

Chemistry

2 answers:

marusya05 [52]3 years ago

3 0

The head of the phospholipids molecule is attracted to water, the tail repels water

photoshop1234 [79]3 years ago

3 0

the head of the phospholipids molecules attracted to water, the tail repels water

You might be interested in

In the reaction below how would adding more of product C affect the equilibrium of the system? A+B arrows both ways C+D

Shalnov [3]

Answer:

1. The reaction will proceed backward, shifting the equilibrium position to the left.

2. The reaction will proceed forward, shifting the equilibrium position to the right.

3. Either add more of the products ( H2O or Cl2) or remove the reactant (HCl or O2)

Explanation:

3 0

3 years ago

Name two metals alloyed with iron to make stainless steel.

kirill115 [55]

For stainless steel different kinds of compositions are used. Based on that different series of stainless steel has been coined.
1. Series 200 - Iron alloyed with <span>chromium, nickel and manganese.
2. Series 300 - It has
a. Stainless Steel 304 - it has composition of 18% chromium and 8% Nickel
b. </span>Stainless Steel 316 - This has 18% chromium and 10% Nickel

Each kind of stainless steel is of different cost and has different applications.

7 0

3 years ago

Suppose that, from measurements in a microscope, you determine that a certain layer of graphene covers an area of 1.50μm2. Conve

irga5000 [103]

One $\mu m$ is $10^{-6}m$ , hence one $\mu m^2$ is $10^{-12}m^2$ , thus $1.50\mu m^2$ is $1.50\cdot10^{-12}\mu m^2$

4 0

3 years ago

Read 2 more answers

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$