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lys-0071 [83]
3 years ago
9

please help 13 points!!! cell wall chloroplasts mitochondria small vacuoles eukaryote cells How would you change the characteris

tics in this list, to describe the cells of organisms in the PLANT kingdom? A) remove mitochondria B) change eukaryote to prokaryote C) change cell wall to cell membrane D) change small vacuole to large central vacuole
Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

A.

Explanation:

It’s just right

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Investigators studying a particular cell response are curious about what type of cell-surface receptor might trigger it. They in
belka [17]

Answer:

An ion channel, more specifically a calcium channel.

Explanation:

The electrical activity of the cells is regulated by ion channels. Calcium channels, also referred as the voltage-gated calcium channels constitute one group of a superfamily of ion channels. A change in voltage across the membrane or small molecules triggers calcium channels to open, allowing calcium to flow into the cell. Inside the cell, calcium acts as a second messenger, it binds to calcium sensitive proteins to induce different responses and support several functions such as muscle contraction, hormone and neurotransmitter secretion, gene regulation, activation of other ion channels, control of action potentials, cell survival, etc.

3 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
3 years ago
what would be the effect on the melting point of a sample if it were not dried completely after filtering the recrystallized sam
jok3333 [9.3K]

The melting point of the sample if it is not dried completely after filtering the recrystallized product will have a broad range and will occur at lower range than the actual value.

What is melting point?

Melting point is the temperature at which the solid form of a given substance changes to the liquid form at atmospheric pressure. It is important because, by using the physical property of a substance the substance can be identified.

The sharp range melting point of the substance indicates the purity of the substance. If the sample is not dried completely after recrystallisation, the melting point will have a broad range.

Therefore, if the sample given is not dried completely, it will be impure and the decreases the melting point of the substance. So the actual melting point of the substance cannot be determined.

To learn more about the melting point click on the given link brainly.com/question/40140

#SPJ4

7 0
1 year ago
Which of the following is an isotope of scandium?<br> 40 X 40 X 4X 2X
Liula [17]
688x




Explanation- Your welcome
6 0
3 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

<u>Step 1:</u> Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

<u>Step 2:</u> The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

<u>Step 3:</u> Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

<u>Step 6:</u> Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

<u>Step 7: </u>Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

<u>Step 8</u>: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0
3 years ago
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