So My friend has started her you know what and she scared to call her mom what should I do I'm sorry if this question is personal anything but I'm kind of confused cuz she's scared to call her mom her mom is gone so she needs some advice the best person who got the best answer will receive a brainly and 50 more points

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:

Answer:

umm

Explanation:

matter makes up everything. everything is matter

the more matter an object has the more mass It has

Answer:

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2.A

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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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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].

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:

helium

Explanation:

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 !

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:

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:

read the story

Explanation:

read it and then you will find the answer

Answer:

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

✔ London dispersion forces

Explanation:

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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Answer:

ask you subject teacher

Explanation:

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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Answer:

Na2S

Explanation:

hope this helps

Answer:

A: Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions. It is likely to react with an element with six valence electrons that “wants” to gain two electrons. This would be an element in group 6, such as oxygen.

Answer:

NaHCO3

Explanation:

The three is small at the bottom at the end

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.

Answer:

This tells you that one mole of sulfur atoms, S , has a mass of 32 g

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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In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.

Let's consider the following balanced redox reaction.

2 Sc + 3 Br₂ ⟶ 2 ScBr₃

We can identify both half-reactions.

Oxidation: 2 Sc ⟶ 2 Sc⁺³ + 6 e⁻

Reduction: 6 e⁻ + 3 Br₂ ⟶ 6 Br⁻

As we can see, 6 electrons are involved in the formation of 2 formula units of ScBr₃. Thus, 3 electrons are involved in the formation of 1 formula unit of ScBr₃.

In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.

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Answer:

1.) 7

2.) 5

3.) stable

4.) 5A

5.) 5

6.) The properties would be more stable to property of phosphorous because same group and valenve electronics and they also have same chemical

Explanation:

The rest i dont know

Answer:

balanced in ACID not BASE

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Answer

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Explanation:

Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)

add H^1+ (acid) to capture the O and make 7 water molecules

Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O

Cr goes from +6 to +3 by gaining 3 e

Hg goes from 0 to +2 by losing 2 e

we need 3 Hg for every 2 Cr

so

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

2 Cr on the right and left

Net 12 positive charges on the right and the left

3 Hg on the right and left

14 H on the right and left

the equation is balanced

we cannot balance the equation in a basic solution with OH^1-

we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid

The balanced molecular chemical equation for the reaction is (NH4)2SO4(aq) + FeCl2(aq) -------> FeSO4(aq) + 2NH4Cl(aq).

The rule for writing a balanced chemical reaction equation is that the number of atom of each element on the right hand side must be the same as the number of atoms of the same element on the left hand side of the reaction equation. This is achieved by performing an atom count.

The balanced molecular chemical equation for the reaction in aqueous solution for ammonium sulfate and iron(II) chloride is;

(NH4)2SO4(aq) + FeCl2(aq) -------> FeSO4(aq) + 2NH4Cl(aq)

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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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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:

what's the question?

Explanation:

I'll help

Answer:

you're welcome