Who wants to simp for me??

by bouncing it

Explanation:

when you bouce a ball it hits the ground because gravity pulls it down

Answer:

the lowest point of the wave is called Trough

Answer:

Trough is the correct answer

Explanation:

The highest part of the wave is called the crest. The lowest part is called the trough.

The number of moles of mercury can you get you collect and process the entire volume of 1250 m3 of the water in the pond is 8.7  × 10^-5 moles.

First, we must convert the concentration of mercury in the pond to molar concentration using the relation;

Mass concentration = Molar concentration × molar mass

Molar concentration = Mass concentration /molar mass

Molar mass of mercury = 201 g/mol

Molar concentration = 14 × 10^-9 g/201 g/mol = 6.97 × 10^-11 M

Volume of solution = 1250 m3 or 1250000 L

Number of moles = concentration × volume

Number of moles =  6.97 × 10^-11 M ×  1250000 L

Number of moles = 8.7  × 10^-5 moles

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Three resonance structures  can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.

Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.

The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.

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Explanation:

since 5g saturates 12.5g of water at 10°c

so......x would saturate 50g of water at 0°c

then you can cross multiply

The heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.

The given parameters:

The heat of fusion of the given sample of the ethyl alcohol converted from solid to liquid phase is calculated as follows;

[tex]H_f = \frac{Q}{m} \\\\H_f = \frac{3.833 \times 10^3\ J}{36.8 \ g} \\\\H_f = 104.16 \ J/g[/tex]

Thus, the heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.

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Answer:

because water is actually a limited natural resource

Explanation:

although most of the earth is made up of water , dasalination plants are super expensive and hard to use. Thus slowly due to severely low rainfall over the years the earth is slowly depleting on levels of fresh water thats why we need to be careful how we use water :D !

This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

[tex]y=-3.5 x + 25\\\\y=-0.52 x + 2[/tex]

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

[tex]-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72[/tex]

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

[tex]y=-3.5 (7.72) + 25\\\\y = 1.84[/tex]

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.

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Answer:

[tex]Molarity\,\,of\,\,the\,\,solution\,\,is\,\,S=0.039M[/tex]

Explanation:

[tex]W=2.18 g\\M=322g\\V=175mL=0.175L\\\\S=\frac{W}{MV} \\=>S=\frac{2.18}{322*0.175} \\So,S=0.039M[/tex]

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Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.

Acids can be strong or weak. This is determined by its tendency to break apart into ions or stay together to form molecules. Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a high number of ions within their solutions.

Weak acids, on the other hand, are those that tend to stay together in the form of molecules and therefore possess very low ion counts in their solutions. The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.

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Answer:

Explanation:

we will do ratio method

Aluminum chloride :  Aluminum

2                               :      2

0.48                          :      x

(cross multiply)

0.48 x 2 / 2 = 0.48 moles of aluminum  

mass = 1 mole of aluminum chloride x moles

mass = 133.33 x 0.48

mass = 63.9984g (round off) = 64g

I hope this helps.

Answer:

LETTER

1. D. PURE SUBSTANCES CAN BE FURTHER BROKEN DOWN INTO SIMPLER SUBSTANCES.

2.A

PO PLEASE MARK A BRAINLIESTS ANG MARK A 5 STAR ANG HEART PO PLEASW

There are different ways of extraction. The best way to begin an extraction to separate the amine from the mixture is to extract with dilute NaOH.

NaOH is known to be the most common compound that is used to convert a carboxylic acid into its more water-soluble ionic carboxylate form.

But if  the mixture has a compound that you want, and that can react with NaOH, another milder base such as sodium bicarbonate is preferably used.

See full question below

A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction in order to separate the carboxylic acid from the mixture?

A) Extract with dilute NaOH

B) Extract with dilute HCl

C) Extract with dichloromethane

D) Extract with water

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The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.

Molar mass of sodium carbonate = 106 g/mol

Molar mass of potassium carbonate = 138 g/mol

Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles

Mass of HNO3 =  0.008903 moles × 63 g/mol = 0.56 g

Mass of sample added = 1.091 g

Mass of sample left over = 0.573 g

Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g

The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.

Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O

106g of Na2CO3 reacts with 126g of HNO3

x g of Na2CO3 reacts with (126 × x/106)g of HNO3

K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O

138 g of K2CO3 reacts with 126 g of HNO3

(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g

Total mass of HNO3 used;

1.19x + 0.47 + 0.91x = 0.56

2.1x + 0.47 = 0.56

2.1x = 0.56 - 0.47

2.1x = 0.09

x =  0.09/2.1

x = 0.0043 g

Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g

Mass percent of K2CO3  = 0.5137 g/ 1.091 g × 100/1 = 47.1%

Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%

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Answer:

read the story

Explanation:

read it and then you will find the answer

Explanation:

Sorry, I don't know, but I can tell you that when an atom, or a body, has the same amount of positive charges (protons) and negative charges (electrons), it is said to be electrically neutral. ... The net charge corresponds to the algebraic sum of all the charges that a body possesses.

Answer: Neutral carbon-12 (or any carbon atom) has 6 electrons with a total negative charge of 6e- orbiting a nucleus with a total positive charge of 6e+, so that the total net charge is zero. The nucleus is made up of 6 protons, each with a positive charge of e+, and 6 neutrons, each with zero charge.

Explanation:

Answer:

umm

Explanation:

matter makes up everything. everything is matter

the more matter an object has the more mass It has

Answer:

helium

Explanation:

Answer:

NaHCO3

Explanation:

The three is small at the bottom at the end

Answer:

A mixture of chlorine gas (Cl2) and fluorine gas (F2):

✔ London dispersion forces

Explanation:

Answer:

from the words below underline six example of rhetorical patterns

Answer:

Approximately [tex]83[/tex] minutes.

Explanation:

Look up the relative atomic mass of [tex]\rm Ag[/tex]: [tex]M({\rm Ag}) = 107.868\; \rm g \cdot mol^{-1}[/tex].[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 1.6987 \times 10^{4}\; \rm C \end{aligned}[/tex]/.

Avogadro's number: [tex]N_A \approx 6.02 \times 10^{23}\; \rm mol^{-1}[/tex].

Elementary charge: [tex]e \approx 1.602 \times 10^{-19}\; \rm C[/tex].

Calculate the quantity of [tex]\rm Ag[/tex] atoms to reduce:

[tex]\begin{aligned}& n({\rm Ag}) \\ &= \frac{m({\rm Ag})}{M({\rm Ag})} \\ &= \frac{19\; \rm g}{107.868\; \rm g \cdot mol^{-1}} \\ & \approx 0.176\; \rm mol\end{aligned}[/tex].

By the equation, it takes one electron to reduce every [tex]\rm Ag[/tex] atom. Thus, the number of electrons required to reduce [tex]0.176\; \rm mol[/tex] of [tex]\rm Ag\![/tex] atoms would be:

[tex]n(e) = n({\rm Ag}) \approx 0.176\; \rm mol[/tex].

[tex]\begin{aligned}N(e) &= n(e) \cdot N_{A}. \\ &\approx 0.176\; \rm mol \times 6.02 \times 10^{23}\; \rm mol^{-1} \\ & \approx 1.06 \times 10^{23}\end{aligned}[/tex].

Calculate the amount of charge (in coulombs) in that many electrons:

[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 16987.1 \; \rm C \end{aligned}[/tex].

A current of [tex]1\; \rm A[/tex] carries a charge of [tex]1\; \rm C[/tex] every second. Thus, the amount of time required for this current to carry that much electron would be:"

[tex]\begin{aligned}t &= \frac{Q}{I} \\ &\approx \frac{16987.1\; \rm C}{3.4\; \rm A} \\ &\approx 83.3\; \rm s \\ &\approx 5.00\times 10^{3}\; \rm s \\ &\approx 83\; \text{minutes} \end{aligned}[/tex].

Answer:

3

Explanation:

Here's the balanced equation;

C2H4 + 3O2 → 2CO2 + 2H2O

C ⇒2                          C ⇒1 x 2 = 2

H ⇒4                          H ⇒2 x 2 = 4

O ⇒3 x 2 = 6             O ⇒ (2 x 2) + (2 x 1) = 6

The coefficient that has been added as the coefficient to O2 is 3.

C2H4 + 3O2 → 2CO2 + 2H2O.

the given equation has 2oxygen atoms on the reactant side and 3 oxygen atoms on the product side

so to make it equal we have to add 3 as coefficient of O2  on the reactant side.

Therefore the correct answer to the equation C2H4 + 3O2 → 2CO2 + 2H2O is 3.

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#SPJ2

Answer:

Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

The reactants are present in equimolar amounts, so there is no excess or limiting reactants.

(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O

(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J

(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O

Explanation:

From the data obtained from the tests, the actual boiling point lies between 34°C and 40°C.

The boiling point of a substance is the temperature at which the pressure of the substance becomes equal to atmospheric pressure. Pure substances have a sharp boiling point while impure substances boil over a temperature range.

In this case, the boiling point of the substance after 10 tests is obtained as 37+/−3°C. This implies that the actual boiling point lies between 34°C and 40°C.

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Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]

Thus, the density of the calcite sample will be:

[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

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Regards!

Answer:

ask you subject teacher

Explanation:

Answer:

you're welcome