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kow [346]
2 years ago
8

Nayla grew skin in a lab by adding cells to a synthetic material. The skin functioned normally for 12 days. Then Nayla separated

some
of the skin cells into cell membranes, cytoplasm, and vacuoles to studythem. WhichofthefollowingwerealiveduringNayla’s experiment?
a. The skin and the cytoplasm
b. The skin and the skin cells
c. The cell membranes and the skin cells
d. The cell membranes and the cytoplasm
Chemistry
1 answer:
iris [78.8K]2 years ago
6 0

Answer:B) The Skin and The Skin cells

Explanation:

Hope this helped

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The electron configuration of an element is 1s22s22p4 Describe what most likely
svetlana [45]

Answer:

The two atoms will contribute and share two electrons each in a double covalent bond, to form a molecule of the element.

Explanation:

Following the rule which says that an octet structure in the valence electrons shell of an atom, when two atoms of an element with the electronic configuration 1s²2s²2p⁴ comeclose to each other, they will share there electrons in covalent bonding to form a molecule, and thus attain stability by achieving an octet structure.

An atom of the element with electronic configuration 1s²2s²2p⁴, has 6 valence electrons and thus, needs two more electrons in order to achieve an octet. This is achieved by sharing two electrons with another atom of the same element which contributes two electrons as well to form a double covalent bond and a molecule of the element. Therefore, each atom would then have an octet structure in their valence shells.

3 0
3 years ago
A football player with a mass of 200 kilograms sprints towards the goal line. Another football player with a mass of 120 kilogra
mihalych1998 [28]

Answer:

oh i love football. i know this. The guy with 200 kilograms because he is heavier o that means more weight going against the guy. Is this American football. cause i love American football  

Explanation:

5 0
2 years ago
Read 2 more answers
Question attached pls help
Bas_tet [7]

Answer:

D sorry it's wrong

Explanation:

4 0
2 years ago
3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 1
RUDIKE [14]

Answer:

Enthalpy change for the dissolution of the unknown solid = 4.6 Kj/mole

Explanation:

Using Q = m x Cs x ΔT ................................(1)

where Q = Amount of heat absorbed

            m = Mass of solution

           Cs = Specific heat capacity of solution

          ΔT = Change in temperature

Given Density of solution = 1.20 g/ml

And volume of solution = 150 ml

Mass of solution = density x volume

                           = 1.2 x 150

                           = 180 g

From equation (1)

              Q = 180 x 4.18 x 19.2 = 14446.08 = 14.4 Kj

So, ΔH  of the dissolution of the unknown solid = \frac{Q}{n} = \frac{14.4}{3.15} = 4.6 kj/mole

7 0
3 years ago
n-Butane fuel (C4H10) is burned with the stoichiometric amount of air. Determine the mass fraction of each product. Also, calcul
tia_tia [17]

Answer:

  1. 0.1852
  2. 0.0947
  3. 0.7201
  4. 3.0345 kg CO_{2} / Kg C_{4} H_{10}
  5. 15.3848 Kg air / kg C_{4} H_{10}

Explanation:

Molar masses of each product are :

Butane = 58 kg /kmol

Oxygen = 32 kg/kmol

Nitrogen = 28 kg/kmol

water = 18 kg/kmol

<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>

= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )

= 176 / 950.32

= 0.1852

<em><u>2) Calculate the mass fraction of water </u></em>

= ( 5 * 18 ) /  (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))

= 90 / 950.32

= 0.0947

<em><u>3) Calculate the mass fraction of Nitrogen </u></em>

= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))

= 684.32 / 950.32

= 0.7201

<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>

Mco2 = ( 4 * 44 ) / 58  = 3.0345 kg CO_{2} / Kg C_{4} H_{10}

<u>5) Mass of Air required per unit of fuel mass burned </u>

Mair = ( 6.5 * 32 + 24.44 *28 ) / 58  = 15.3848 Kg air / kg C_{4} H_{10}

5 0
2 years ago
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