Q1.

Blizzard [7]

Your answer would be 5.

4 0

2 years ago

The ________ molecules which form the bilayer region of the cell membrane have hydrophilic regions on the outer surface and hydr

Goshia [24]

Answer:

The answer is: phospholipid molecules

Explanation:

The plasma membrane of a cell is consists of a lipid bilayer. This lipid bilayer, also known as the phospholipid bilayer, is a polar membrane composed of two layers of lipid molecules, usually amphipathic phospholipid molecules.

The amphipathic phospholipid molecules have a hydrophilic phosphate head on the exterior and a hydrophobic tail consisting of fatty acid chain on the interior of the membrane.

8 0

2 years ago

What natural and human made features can a map show? Give two examples of each.

Bess [88]

Natural: Land mounds, Rivers
Human made: Country borders, cities

3 0

3 years ago

A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm

VARVARA [1.3K]

Answer: The nuclear equations for the given process is written below.

Explanation:

The chemical equation for the bombardment of neutron to U-238 isotope follows:

$_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}$

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

The chemical equation for the first beta decay process of $_{92}^{239}\textrm{U}$ follows:

$_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta$

The chemical equation for the second beta decay process of $_{93}^{239}\textrm{Np}$ follows:

$_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta$

Hence, the nuclear equations for the given process is written above.

6 0

2 years ago

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the balanced equation 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago