Help i dont get it

Answer:

Because of the two different parts of the molecule, a soap molecule is soluble in water and at the same time can dissolve fats. The fats are attracted to the non-polar tail part of the soap while the polar head makes the whole complex.

Explanation:

hope it helps

Answer:

Explanation:

Consider the general chemical reaction

                                                       [tex]\mathrm{A} \ \overset{k}{\longrightarrow} \ \mathrm{product}[/tex]  .

If [A] is the concentration of A (reactant) at any time t and n is the reaction order for the whole equation, the rate is then related to the concentration of reactant A with the following differential form of equation

                                               [tex]Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^{n}[/tex]  .

where k is the rate constant.

*Note that the differential term [tex]\displaystyle\frac{d[\mathrm{A}]}{dt}[/tex] has a negative sign to denote that the concentration of A is decreasing over time t.

Since the chemical reaction between tert-butyl bromide and water is given to be a first-order reaction, hence n = 1, and the resulting differential equation becomes

                                        [tex]Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^{1} \ = \ k[\mathrm{A}][/tex]

To solve this first-order linear homogenous differential equation, the method of separation of variables can be used.

[tex]\-\hspace{1cm} \displaystyle\frac{d[\mathrm{A}]}{dt} \ = -\ k[\mathrm{A}] \\ \\ \-\hspace{0.5cm} \displaystyle\frac{1}{[\mathrm{A}]} \, d[\mathrm{A}] \ = -\ k \, dt \\ \\ \int\ {\displaystyle\frac{1}{[\mathrm{A}]} \, d[\mathrm{A}] \ = \ -\int {k} \, {dt}[/tex]

         [tex]\ln{[\mathrm{A}]} \ = \ -kt \ + \ C \\ \\ \-\hspace{0.45cm} $[A]$ \ = \ e^{-kt \ + \ C} \\ \\ \-\hspace{0.45cm} $[A]$ \ = \ e^{-kt}e^{C} \ \ \ \ \ \ \ \ \ (e^{a \ + \ b} \ = \ e^{a}e^{b} \ \ \ \mathrm{by \ the \ law \ of \ indices})[/tex]

Since the term [tex]e^{C}[/tex] is a constant, let [tex]\alpha \ = \ e^{C}[/tex], hence [tex][\mathrm{A}] \ = \ \alpha e^{-kt}[/tex] or [tex]C \ = \ \alpha e^{-kt}[/tex] according to the question. Given that the initial concentration (t = 0) of tert-butyl bromide is 0.2 mol/L and k = 0.0537[tex]\mathrm{h^{-1}}[/tex], so

[tex]0.2 \ = \ \alpha e^{-0.0537 \ \times \ 0} \\ \\ 0.2 \ = \ \alpha \ \ \ \ \ (e^{0} \ = \ 1)[/tex]

Therefore, the rate equation is [tex]C \ = \ 0.2e^{-0.0537t}[/tex].

The concentration of tert-butyl alcohol after 1 hour is

[tex]C \ = \ 0.2e^{-0.0537 \ \times \ 1} \\ \\ C \ = \ 0.2e^{-0.0537} \\ \\ C \ = \ 0.190 \ \mathrm{mol/L} \ \ \ (3 \ \ \mathrm{s.f.})[/tex]

Answer:

Answer: 0.66 hours

Explanation:

[tex]{ \rm{time = \frac{distance}{speed} }} \\ \\ { \rm{time = \frac{(950.5 \times 1000) \: m}{400 \: m {s}^{ - 1} } }} \\ \\ { \rm{time = {2376.25 \: s}}}[/tex]

• Remember 1 second accounts for 1/3600 hours

[tex]{ \rm{time = \frac{2376.25}{3600} }} \\ \\ { \rm{time = 0.66 \: hours}}[/tex]

The time taken by object  traveling at a speed of 400 m/s to reach 950.5 km distance is 0.66 hr or 39.6 min.

What is average velocity ?

Average velocity is given by the total distance covered by the object in total given time. SI unit of velocity is m/s.

Here in this question it is given that the object is travelling with speed v of 400 m/s and it is to find the time t to reach 950.5 km distance d.

it is known that 1 km = 1000 m

that is 950.5 km = 950500 m

Now, using the relation of speed-distance to find time t to reach 950.5 km distance :

[tex]\begin{aligned}v&=\frac{d}{t}\\t&=\frac{d}{v}\\&=\frac{950500}{400}\\&=2376.25 \text{ s}\end{aligned}[/tex]

Now converting 2376.25 sec into hrs by dividing it by 3600 we get :

2376.25/3600 = 0.66 hr or 39.6 min.

Therefore, the time taken by object to reach 950.5 km distance is 0.66 hr or 39.6 min.

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Answer:

i did not find the amswer sorry

Answer:

A) Carbon dioxide and Oxygen

Explanation:

Percentage of gases in the Planet's atmosphere:

Percentage of gases in the Earth's atmosphere:

___________________

On observing the percentage composition of the atmospheres of the two Planets, we get:

___________________

Answer:

What's asked is the name of the gases that are in higher amounts in the atmosphere of the newly discover planet

Therefore,

Answer: D.

Explanation:

Asexual reproduction requires only one parent, making it more efficient and take a shorter amount of time than sexual reproduction.

Answer:

true

Explanation:

Because the changes in state of matter would be considered

The elements in the lanthanide series are similar to gadolinium.

The periodic table is arranged in groups and periods. Elements in the same group have similar properties. At the bottom row of the periodic table lies a group of elements that have unique properties called the lanthanides. This is where Gadolinium belongs.

The elements in the lanthanide series such as europium and terbium are similar to gadolinium. The lanthanides are similar in chemical reactivity and appearance. However, the density of the metals differ because it increases with increase in atomic number.

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The maximum amount of a substance that can be dissolved in a solvent at a given temperature is referred to as its solubility.

Sucrose has a solubility of 320 g/100 g H2O at 70°C.

Sucrose has a solubility of 320 g/100 g H2O at 70°C.

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Answer:

A carbon dioxide and oxygen

Answer:

C 7.5 Moles of MgSO4

Explanation:

The method of propagation of the given organisms is shown below:

1. Mongo ---> seeds

2. Moss ---> spores

3. Ferns ---> stem

4. Pine Tree ---> cones

5. Guava ---> seeds

6. Mango ---> seeds

7. Coconut ---> seeds

8. Katakataka ---> leaves

9. Potato ---> stem

10. Okra ---> seeds

Repr0duction is the process by which living organisms produce young ones of their kind.

Repr0duction can either involve two parents through the joining of male and female gametes to produce a zygote i.e. seeds in plants and embryo in animals or can involve only one parent and occurs through vegetative parts such as roots, stem, or leaves.

The table below gives organisms and their propagation parts:

1. Mongo ---> seed

2. Moss ---> spores

3. Ferns ---> stem

4. Pine Tree ---> cones

5. Guava ---> seeds

6. Mango ---> seeds

7. Coconut ---> seeds

8. Katakataka ---> leaves

9. Potato ---> stem

10. Okra ---> seeds

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Answer:

true or false

a. very little oxygen and nitrogen are found in air. false

b. oxygen is the main cause of the formation of rust in iron.   false

c.  impure air is odourless true

.d.  plants take in oxygen gas during respiration. true

e. air contains water vapour. ​true

Explanation:

Answer:

It's false

Explanation:

Answer:

to identify the element by using flame tests

heterotrophs cannot create their own food and rely on others for food

Answer:

D

Explanation:

The valence electrons are the outer electrons and must be located in the outermost shell. In this case, D.

The probability density function is nonnegative everywhere, and its integral over the entire space is equal to the terms "probability distribution function" and "probability function" .

In probability theory, a probability density function (PDF), or density of a continuous random variable, is a function whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative likelihood that the value of the random variable would be equal to that sample.

Probability density is the probability per unit length, in other words, while the absolute likelihood for a continuous random variable to take on any particular value is 0 (since there is an infinite set of possible values to begin with), the value of the PDF at two different samples can be used to infer, in any particular draw of the random variable, how much more likely it is that the random variable would be close to one sample compared to the other sample.

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Explanation:

What do you mean by 'how many oxygen'?

oxygen atoms? or oxygen percentage?

if oxygen atoms is 12 atoms

Answer:

Gravity will always be pulling everything towards Earth's center at an acceleration of 9.81 m/sec².

Answer: it wound'nt let me type the equations out so i did it on g.o.o.gl.e docs :')

Explanation:

hope this helps <3

have a great day/night/noon

:)

Answer:

Example of ionic bonds— Sodium chloride and calcium oxide. Example of covalent bond— Water, carbon

Explanation:

ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound. Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom.

A covalent bond consists of the mutual sharing of one or more pairs of electrons between two atoms. These electrons are simultaneously attracted by the two atomic nuclei. A covalent bond forms when the difference between the electronegativities of two atoms is too small for an electron transfer to occur to form ions.

Answer:

child stage.

Explanation:

they are more vubnlerabke at that stage

Answer:

A- Dry with low humidity

Explanation:

The hydrosphere includes the water which is on the surface of the earth, below the ground and in the air.  

On analyzing the climatic conditions in the tropics due to the interaction of the atmosphere and hydrosphere, the presence of water droplets in tropics is less resulting in less humidity and, thus the climatic condition is dry.

Therefore the climatic condition that is typically found in tropics due to interaction of the atmosphere and hydrosphere is dry with low humidity.

Hope this helps!

Explanation:

Are you asking if that is the formula for caprylic acid? The answer is yes. CH₃(CH₂)₆COOH is the correct formula.

Caprylic acid, also known as C8 or octanoic acid, is a medium chain triglyceride (MCT) primarily found in coconut oil. It's the shortest of all the MCTs, making it the easiest for the body to break down into fuel. Once it is consumed, C8 is instantly digested and converted into ketones.

Answer:

0.16 L

Explanation:

Step 1: Write the balanced equation

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Step 2: Calculate the moles of CO₂ produced from 0.012 moles of O₂

The molar ratio of O₂ to CO₂ is 5:3. The moles of CO₂ produced are 3/5 × 0.012 mol = 0.0072 mol.

Step 3: Calculate the volume occupied by 0.0072 moles of CO₂ at standard temperature and pressure

At STP, 1 mole of CO₂ occupies 22.4 L.

0.0072 mol × 22.4 L/1 mol = 0.16 L

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Answer:

i the answer to your question should be C

Explanation:

Hope this helps!

The atomic mass of an element related to a mole in grams, is the mass of one mole of atoms. Therefore, option C is correct.

An atom's mass is its atomic mass. Although the kilogramme is the SI measure of mass, the unified atomic mass unit, or dalton, is a common way to express atomic mass. An unbound carbon-12 atom in its ground state has a mass of 112 of a Da.

A mass that is one-twelfth the mass of an atom of carbon-12 is known as an atomic mass unit. Any isotope of any element's mass is represented in terms of the carbon-12 reference standard.

The size of an atom is indicated by its atomic mass. Technically speaking, an atom's mass is the total weight of its protons, neutrons, and electrons.

Thus, option C is correct.

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If there is direct variation and y=75 when x=25,when y=48 X = 16. Therefore, option A is correct.

link between two variables that can be described mathematically by an equation where one variable equals a constant multiplied by the other. For instance, the constant of variation is k = = 3 if y varies straight as x and y = 6 when x = 2. Consequently, y = 3x is the equation that describes this direct variation.

When x is not equal to zero, an equation of the form y = kx describes the linear function known as direct variation. When x is not equal to zero and k is a nonzero real number constant, the equation of the form xy = k describes the nonlinear function known as inverse variation.

x = ky

where k is constant

25 = k × 75

25 / 75 = k

k = 1 / 3

now when y = 48 then substitute the values

x = ky

x = 1 ÷ 3 × 48

x = 16

Thus, option A is correct.

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