Oxygen and hydrogen form the polyatomic hydroxide ion. What is its charge?

Answer:

Explanation:

We can calculate the Rf values by using the following formula:

Rf = Distance from the baseline / Solvent front distance

With that in mind we now proceed to calculate the Rf value for both components:

Explanation:

example is copper iron...........

Answer:

See explanation

Explanation:

When using a separating funnel to separate a mixture of substances, we expect that the system will consist of an organic phase and an aqueous phase. The organic phase is usually found on top while the aqueous phase is found below.

When naphthalene, benzoic acid and ethyl 4-aminobenzonate mixed in ether solution is separated using HCl, the ethyl 4-aminobenzonate is found in the lower aqueous layer.

This is because, the ethyl 4-aminobenzonate can react with HCl to form an ionic hydrochloride salt which dissolves in the lower aqueous layer.

Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders

Answer:

Excitation of Carbon atom in CH, involves promotion of one is electron to the empty 2p-orbital

Answer:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Double replacement reaction.

It is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Explanation:

Hello there!

In this case, according to the given information, it turns possible for us to solve this problem by firstly considering that this reaction occurs between potassium iodide and lead (II) nitrate to yield potassium nitrate and lead (II) iodide which is clearly not balanced since we have one iodine atom on the reactants and two on the products, that is why the balance implies the placement of a coefficient of 2 in front of both KI and KNO3 as shown below:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Thus, we infer this is a double replacement reaction due to the exchange of both cations, K and Pb with both anions, I and NO3. Moreover, we can tell this balanced reaction is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Regards!

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

Answer:

164

Explanation:

40+2(14)+6(16)

40+28+96=164

The formula mass of a compound is the sum of  masses individual elements in that compound. The formula mass calcium nitrate  (Ca (NO₃)₂) is 164 g/mol.

Formula mass of a compound is calculated from the empirical formula  of the compound. The mass of one mole of an element is called its atomic mass. The mass of an element in a compound is its subscripts times the atomic mass.

The relative formula mass of a compound is the sum of the relative masses of each element present in the compound multiplied with their number of atoms.

For the compound Ca (NO₃)₂, there are one calcium atom and two nitrate groups.

atomic mass of Ca = 40 g/mol

mass of oxygen = 16 g/mol

mass of 3  O = 48 g.

mass of N = 14 g/mol

Then, mass of  two nitrate groups = 2× (48 + 14) =124 g

total mass = 124 g + 40 = 164 g/mol

Therefore, the formula mass of the compound Ca (NO₃)₂ is 164 g/mol.

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Answer:

its letter b

Explanation:

it represents the spectrum of stars

Answer:

[tex]Q=292240.8J=292.2kJ[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to use the general heat equation:

[tex]Q=mC(T_2-T_1)[/tex]

For us to plug the given mass, specific heat and temperature change to obtain the required heat:

[tex]Q=120g*32.91\frac{J}{g*K} (98\°C-24\°C)\\\\Q=292240.8J=292.2kJ[/tex]

Regards!

Answer:

0.030 mole

Explanation:

Mass of crucible + lid = 23.422 g

Mass of crucible + lid + compound = 24.746 g

Mass of crucible + lid + compound - water = 24.213

Mass of water = Mass of crucible + lid + compound + heat

       = 24.746 - 24.213

                = 0.533 g

Mole of water in the hydrated compound = mass of water in the compound/molar mass of water

    = 0.533/18

         = 0.0296 mole = 0.030 mole

Explanation:

here's the answer to your question

Answer:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Explanation:

The strongest acid is the one that has the higher Ka. Now, the weakest conjugate base is the conjugate base of the strongest acid and vice versa:

In the problem, we have only conjugate bases, as the HCN is the weakest acid, the strongest conjugate base is NaCN, then KClO and as last C2H5NH3Br and NaI (The conjugate base of a strong acid, HI).

The strongest base has the higher pH, that means. Thus, the rank in order of increasing pH is:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Answer:

0.000731 grams aluminium sulfate

46.0 mols ammonia

Explanation:

ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol

[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]

NH3 has a molar mass of 17.031 g/mol

[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]

Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]

we have to find the 0.250 moles of aluminum sulphate.

[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]

[tex]\\\\\\[/tex]

Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]

We have to find 2.70 grams of ammonia

[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]

Answer:

Following are the chemical equation to the given question:

Explanation:

The Electrode is a silver film that is covered with such a thin coating of silver chloride, either by dipping its wire directly into silver-molten chloride, plating the wire using hydrogen peroxide, or oxidation silver in a chloride. In the given silver electrode, this anode acts as a cathode and thus reduces.

Half of the response reduction: [tex]Ag^+(aq)+e^-\rightarrow Ag(s)[/tex]

Half-effect oxidation:  [tex]Pb(s)\rightarrow Pb^{2+}(aq)+2e^-[/tex]

Complete reaction:  [tex]Pb(s)+2Ag^+(aq) \rightarrow Pb^{2+}(aq)+2Ag(s)[/tex]

Answer:

irreversible.

I hope this will help you

Answer:

A.  The entropy of the universe is increasing.(always)

Explanation:

The Second Law of Thermodynamics states that Entropy cannot decrease, because it keeps increasing and increasing and increasing. It will always stay on the increasing side.

Hope this helps!

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Answer:

2Ag + H2O -----> Ag2O + 2H

Explanation:

2Ag + H2O -----> Ag2O + 2H is the equation of the reaction between metal and steam. Silver reacts with water (steam) forming silver oxide and hydrogen gas. When the metals react with steam it produces the solid metal oxide and hydrogen gas. On the surface o metals, a protective layer of aluminium oxide is formed that keeps water away from the metal so we can say that silver oxide and hydrogen are formed from the reaction of silver metal and steam.

Answer:

R.F.M of Iron (II) oxide :

[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]

Moles :

[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]

Molecules :

[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]

The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.

Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.

The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.

Given, the mass of the iron oxide = 35.2 ×10⁻²³ g

The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol

71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³

35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)

The number of molecules of FeO in a given mass = 2.95 molecules

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Answer:

the constant temperature will be 12 .F bec in places it is cold

Answer: Hello your question is poorly written attached below is the complete question

answer:

450 kg

Explanation:

mass of product formed = mass of reactants that reacted

hence :

mass of compound C that can be formed and shipped

= mass of A  +  mass of B

= 200 kg + 250 kg = 450 kg ( theoretical mass of compound formed )

Answer:

1) 2.69 * 10²³ PBr₅

2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁

Explanation:

Question 1)

We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.

Phosphorus pentabromide is given by PBr₅.

To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.

To convert from grams to moles, we will find the molar mass of PBr₅.

Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:

[tex](30.974)+5(79.904) = 430.494\text{ g/mol}[/tex]

And since we want to convert from grams to moles, we can write the following ratio:

[tex]\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}[/tex]

Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.

To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.

So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:

[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]

In combination, starting with 192 grams of PBr₅, we will acquire:

[tex]\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]

Cancel like units:

[tex]\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}[/tex]

Multiply. Hence:

[tex]=2.6858...\times 10^{23}\text{ PBr$_5$}[/tex]

Since the final answer should have three significant digits, our final answer is:

[tex]= 2.69\times 10^{23} \text{ PBr$_5$}[/tex]

So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.

Question 2)

We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.

3.42 kilograms is equivalent to 3420 grams of table sugar.

Again, we can convert from grams to moles, and then from moles to molecules.

First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:

[tex]12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}[/tex]

To cancel units, we can write our ratio as:

[tex]\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}[/tex]

With grams in the denominator.

And by definition:

[tex]\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]

Combining the two ratios and the starting value, we acquire:

[tex]3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]

Cancel like units:

[tex]=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}[/tex]

Multiply:

[tex]\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]

Rewrite:

[tex]\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]

The resulting answer should have three significant digits. Hence:

[tex]=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}[/tex]

So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.

Answer:

2.69×10²³ molecules of PBr₅

6.02×10²⁴ molecules of C₁₂H₂₂O₁₁

Explanation:

To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.

[tex]192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules[/tex]

Now, we know that there are about 2.69×10²³ molecules of PBr₅.

To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.

[tex]3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules[/tex]

Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.

Answer:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Explanation:

Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Answer:

absorption, fluorescence = phosphorescence

Explanation:

Given a particular fluorophore, the wavelength of absorption of the fluorophore is always shorter. Both fluorescence and phosphorescence are kinds of photoluminescence.

Recall that both fluorescence and phosphorescence occur at a longer wavelength. The difference between the two is only in the time taken during the process. While fluorescence takes a shorter time to occur, phosphorescence takes a longer time to occur.

The major difference between fluorescence and phosphorescence is that change of spin occurs during phosphorescence but not fluorescence.

Answer:

3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)

Explanation:

Let's consider the balanced chemical equation that takes place when sulfuric acid solution and solid aluminum hydroxide react to form aluminum sulfate and water. This is a neutralization equation.

3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)

Answer:

The chlorine element belongs to group 17 because it has 7 valence electron . Its valency is 1 . It can gain one electron from any other atom to become stable. This means that it can never form a double or triple bond.

Answer:

1360kJ are evolved

Explanation:

When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.

To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:

Moles H2 -Molar mass: 2g/mol-

90g H2 * (1mol / 2g) = 45 moles

Moles O2 -Molar mass: 32g/mol-

90g * (1mol / 32g) = 2.81moles

For a complete reaction of 2.81 moles of O2 are needed:

2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2

As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.

As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:

2.81 moles O2 * (241.8kJ / 1/2moles O2) =

The quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is

The equation for the combustion of hydrogen is given as:

[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]

Recall that:

Since the mass of 180 g is equally shared by H₂ and O₂, then:

The number of moles of the reactant can be determined as follows:

For H₂:

[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]

no of moles = 44.6 mol

For O₂:

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]

no of moles = 2.8 mol

Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:

If 1/2 moles of O₂ produces -241.8 kJ/mol of water;

Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:

[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]

[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]

= - 1358.91 kJ

≅ - 1360 kJ

Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ

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