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4 years ago

What part of a cell has a function that is similar to your skin's? What is that function? .

Chemistry

1 answer:

jeyben [28]4 years ago

4 0

Cell membrane

Explanation:

The cell membrane is the part of a cell that has similar functions to the skin.

The skin is a protective covering round the body. It also helps in maintaining a dynamic balance between the environment and the body.

Cell membrane is found in both plants and animals. It is made up of rich phospholipid layers that protects the internal component of the cell. It also allows for selective permeability.

learn more:

Eukaryotic cells brainly.com/question/4493579

Cell organelles brainly.com/question/1494707

#learnwithBrainly

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Will give brainliest and extra points please help ASAP

Viefleur [7K]

The Correct answer to this question is translation

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3 years ago

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What is the molarity of a .25 liter solution containing 52 grams of HBr?

Phantasy [73]

Molar mass HBr = 80.91 g/mol

number of moles:

mass of solute / molar mass

52 / 80.91 => 0.642 mols

Volume = 0.25 L

M = n / V

M = 0.642 / 0.25

M = 2.568 mol/L

hope this helps!

4 0

4 years ago

At what temperature (in K) will 3.5 moles of gas occupy 2.7 L at 1.5 atm? HELP PLEASE!

Vesnalui [34]

3.5 moles of a gas will occupy 2.7 L at 1.5 atm at a temperature of 14.1K

IDEAL GAS LAW:

The temperature of a gas can be calculated using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)

According to this question, P = 1.5atm, V = 2.7L, n = 3.5moles, T = ?

1.5 × 2.7 = 3.5 × 0.0821 × T

4.05 = 0.28735T

T = 4.05 ÷ 0.28735

T = 14.1K

Therefore, 3.5 moles of a gas will occupy 2.7 L at 1.5 atm at a temperature of 14.1K

Learn more at: brainly.com/question/13821925?referrer=searchResults

6 0

3 years ago

Which of the following is a cycloalkane?

hodyreva [135]

Answer:

Explanation:

7 0

3 years ago

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The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

<u>Answer:</u> The moles of $CO_2$ added to the system is 7.13 moles

<u>Explanation:</u>

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

4 0

3 years ago