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ivanzaharov [21]
3 years ago
15

What part of a cell regulates the movement of materials in and out of the cell

Chemistry
2 answers:
goblinko [34]3 years ago
4 0
I. plasma membrane - It surrounds the cell and has three major functions: ... It regulates the movement of substances into and out of the cell. It is referred to as being selectively permeable because it allows or prevents certain substances to enter and leave the cell.
storchak [24]3 years ago
3 0
The cell membrane controls what goes in and out of the cell. It is contained in An ANIMAL, PLANT and BACTERIA cell. For more information go to the following website: https://www.bbc.com/bitesize/guides/zpqpqhv/revision/8
You might be interested in
Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol
max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

  • ΔG = ΔH - TΔS

On this note, it follows that;

  • ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

  • ΔG = 14.6 - 71.52
  • ΔG = -56.92KJ

Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

Read more on Gibb's free energy;

brainly.com/question/13765848

5 0
2 years ago
Read 2 more answers
Sort these species into isoelectronic groups. It doesn\'t matter which group goes in which box, so long as the correct species a
OleMash [197]

Answer : The isoelectronic groups are:

He,Be^{2+},Li^{+}

N^{3-},Mg^{2+}

S^{2-},Ti^{4+},K^{+}

Explanation :

Isoelectronic : It is defined as the compound or molecule having the same number of electrons and the same number of electronic structure.

  • The element is helium. The number of electrons are 2.
  • The element is beryllium. The number of electrons are 4. The number of electrons in Be^{2+} = 4 - 2 = 2
  • The element is lithium. The number of electrons are 3. The number of electrons in Li^{+} = 3 - 1 = 2
  • The element is nitrogen. The number of electrons are 7. The number of electrons in N^{3-} = 7 + 3 = 10
  • The element is neon. The number of electrons are 10.
  • The element is sulfur. The number of electrons are 16. The number of electrons in S^{2-} = 16 + 2 = 18
  • The element is magnesium. The number of electrons are 12. The number of electrons in Mg^{2+} = 12 - 2 = 10
  • The element is titanium. The number of electrons are 22. The number of electrons in Ti^{4+} = 22 - 4 = 18
  • The element is potassium. The number of electrons are 19. The number of electrons in K^{+} = 19 - 1 = 18

The isoelectronic groups are:

He,Be^{2+},Li^{+}

N^{3-},Mg^{2+}

S^{2-},Ti^{4+},K^{+}

7 0
3 years ago
Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di
pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

\frac{1}{2}\times 1.046 mol=0.523 mol calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

\frac{2}{1}\times 0.1908 mol=0.3816 molof dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0
3 years ago
If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your
Drupady [299]

N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

<em>N₂+3H₂⇒2NH₃</em>

mol N₂(MW=28 g/mol) :

\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571

mol H₂(MW= 2 g/mol) :

\tt mol=\dfrac{100}{2}=50

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

\tt \dfrac{3.571}{1}\div \dfrac{50}{3}=3.571\div 16.6

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant

5 0
3 years ago
Electric power is measured in watts (1 W = 1 J/s). About 95% of the power output of an incandescent bulb is converted to heat an
nlexa [21]

Answer:

9.430 * 10¹⁷ protons per second whill shine on the book from a 62 W bulb

Explanation:

To answer this question, first let's calculate the energy of a single photon with a wavelength (λ) of 504 nm:

E = hc/λ

Where h is Planck's constant (6.626*10⁻³⁴ J·s) and c is the speed of light (3*10⁸ m/s).

E = 6.626*10⁻³⁴ J·s * 3*10⁸ m/s ÷ (504*10⁻⁹m) = 3.944 * 10⁻¹⁹ J.

So now we can make the equivalency for this problem, that

<u>1 proton =  3.944 * 10⁻¹⁹ J</u>

Now we convert watts from J/s to proton/s:

1 \frac{J}{s}*\frac{1proton}{3.944*10^{-19}J}=2.535*10^{18} \frac{proton}{s} = 1 W

Solving the problem, a 62 W bulb converts 5% of its output into light, so:

  • 62 * 5/100 = 3.1 W

3.1 watts are equal to [ 2.535*10¹⁸ proton/s * 3.1 ] = 7.858 * 10¹⁸ proton/s

Of those protons per second, 12% will shine on the chemistry textbook, thus:

7.858 * 10¹⁸ proton/s * 12/100 = 9.430 * 10¹⁷ protons/s

8 0
3 years ago
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