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3 years ago

What does the nucleus of the cell control?

Chemistry

1 answer:

dusya [7]3 years ago

7 0

Answer:

Image result for What does the nucleus of the cell control?

The nucleus controls and regulates the activities of the cell (e.g., growth and metabolism) and carries the genes, structures that contain the hereditary information. Nucleoli are small bodies often seen within the nucleus.

Explanation:

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Who was the first scientist to propose that an object could emit only certain amounts of energy?

Liono4ka [1.6K]

Answer:

Option A), Planck.

Explanation:

The first scientist to propose that an object could emit only certain amounts of energy was Max Planck.

Incandescent (very hot) objects emit different color of lights, i.e. different wavelengths, but by 1900 it was believed that the light (energy) was emitted in a continuous way.

Since the wave model did not explain the emission of such specific wavelengths, when the German physicist Max Planck (1858 - 1947) studied the light emitted by incandescent objects, he proposed that the objects (the matter) could gain or lose energy only in small specific amounts or packages of energy called quanta (plural of quantum).

Planck, later, established the mathematical relation between the energy of a quantum of light with the frequency of such the radiation (light emitted). This is the formula:

E = hν.

Where, E is the energy, h is Planck's constant, and ν is the frequency.

5 0

3 years ago

The man is being warmed by? radiation convention conduction

Lubov Fominskaja [6]

Answer:

Comvection

Explanation:

The movement causes the density to rise and therefore transfer more heat

6 0

3 years ago

Figure

Artyom0805 [142]

Answer:

Explanation:

7 0

3 years ago

Hydrogen peroxide is a chemical used to clean bacteria out of cuts. It does this by creating germ-eating oxygen in this decompos

weeeeeb [17]

Answer:

Coefficient's are 2, 2, 1

Explanation:

We are given;

Unbalanced equation;

H₂O₂ → H₂O + O₂

We are required to determine the suitable coefficients that would balance the equation.

We need to know that for an equation to obey the law of conservation of mass it has to be balanced.
Balancing involves making sure that the number of atoms of each element are equal on both sides of the equation.

In this case;

To balance the equation we put the coefficients 2, 2, 1 respectively

Therefore;

The balanced equation is;

2H₂O₂ → 2H₂O + O₂

Thus, Suitable coefficients are, 2, 2, 1

5 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago