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larisa [96]
3 years ago
6

What does the epidermis do for the skin?

Chemistry
1 answer:
vampirchik [111]3 years ago
8 0

Addie3137,

The epidermis is the outer layer of the skin. It's function is too "make a skin waterproof it acts as if its a barrier, and it helps give us skin tone." Epidermis also helps us have homeostasis which is how the body keeps cool or warm. Epidermis contains sweat glands which produce sweat once the body reaches it's limit when exposed to heat, and sweating helps keep the body cool.

Hope this helps!

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Shiny; ductile; solid at room temperature which of the elements best fits the description? hint: look at your periodic table.
avanturin [10]
Take the attached picture of a periodic table as a guide. You are finding for a solid metal. Therefore, streamline your choices by looking at elements written in black bold letters, because they are all solid. Next, if you look at the center, the legend for metals are colors in orange, yellow, flesh, lavender, pink, and cyan blue. These region would be your choices. Next, you want to find a metal that is shiny and ductile. The shiny appearance is a common characteristic of luster by materials. Ductility is the ability of a metal to stretch when under tensile stress. These properties are best exhibited by metals in the transitions metals colored in pink. Therefore, the answer to your question would be any of the metal in the pink area. Examples are Titanium, Chromium, Gold, Silver, Platinum, Tungsten, etc.

6 0
3 years ago
An astronaut in her spacesuit weighs 300 lb on earth. What would her weight be on the moon?
Phantasy [73]
She would be 150 on the moon because the moon has half the mass of earth
6 0
3 years ago
Read 2 more answers
Pls help ill mark as brainliest :)
Alina [70]

Answer:

I think its a double reaction

Explanation:

4 0
3 years ago
What volume would a sample of gas occupy in LITERS at 0.985 atmospheres and a volume of 3.65 liters if the pressure were raised
musickatia [10]

Answer:

3.18 L

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 0.985 atm
  • Initial volume (V₁): 3.65 L
  • Final pressure (P₂): 861.0 mmHg
  • Final volume (V₂): ?

Step 2: Convert P₁ to mmHg

We will use the conversion factor 1 atm = 760 mmHg.

0.985 atm × 760 mmHg/1 atm = 749 mmHg

Step 3: Calculate the final volume of the gas

Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L

7 0
3 years ago
It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis
Norma-Jean [14]

Answer:

a)

The overall  balanced combustion  reaction is written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

(F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = 23.562

b)

the higher heating values (HHV)_f per unit mass of LPG = 49.9876 MJ/kg

the lower heating values (LHV)_f per unit mass of LPG = 46.4933 MJ/kg

Explanation:

a)

The stoichiometric equation can be expressed as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2  \ + \ bH_2O \ + \ cN_2

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

x = \frac{2a+b}{2} \\ \\ x =  \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95

Equating the coefficient of Nitrogen

c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612

Therefore, The overall  balanced combustion  reaction can now be written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

Now;  To determine the stoichiometric F/A and A/F ratios; we have:

(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\  (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\  (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562

b)

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8}  = 0.7 \\ \\ \\  m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06  \\  \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6}  = 0.24

Finally; calculating the higher heating values (HHV)_f per unit mass of LPG; we have:

(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg

calculating the lower heating values (LHV)_f per unit mass of LPG; we have:

(LHV)_f = (HHV)_f - \delta H_w \\ \\  (LHV)_f = (HHV)_f  - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f   = 49.9876 \ MJ/kg -  [\frac{3.8*18}{44.2}*2.258 \ MJ/kg]  \\ \\ (LHV)_f = 46.4933 \ M/kg

7 0
3 years ago
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