the number of electrons present in 4.2 gram of methane is​

Answer:

Percentage yield = 38.37%

Explanation:

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃

2 moles of Al contains = (2 × 27) = 54g of Al

1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃

54g of Al produces 342.15g of Al₂(SO₄)₃,

46g of Al will produce x g of Al₂(SO₄)₃

Solve for x

X = (46 × 342.15) / 54

X = 15738.9 / 54

X = 281.46g

Theoretical yield of Al₂(SO₄)₃ = 281.46g

Percentage yield of a substance = (actual yield / theoretical yield) × 100

Actual yield = 108g

Theoretical yield = 281.46

% yield = (108 / 281.46) × 100

% yield = 0.3837 × 100

%yield = 38.37

The percentage yield of Al₂(SO₄)₃ is 38.37

Answer:  Percentage yield = 38.37%

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂  from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃ .

2 moles of Al contains = (2 × 27) = 54g of Al

1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃

54g of Al produces 342.15g of Al₂(SO₄)₃,

46g of Al will produce x g of Al₂(SO₄)₃

Solve for x  

X = (46 × 342.15) / 54

X = 15738.9 / 54

X = 281.46g

Theoretical yield of Al₂(SO₄)₃ = 281.46g

Percentage yield of a substance = (actual yield / theoretical yield) × 100

Actual yield = 108g

Theoretical yield = 281.46

% yield = (108 / 281.46) × 100

% yield = 0.3837 × 100

%yield = 38.37

The percentage yield of Al₂(SO₄)₃ is 38.37

Answer:

CF₂O

Explanation:

To determine how the molecule will be made think about the valence number of each element in molecule:

 F= -1 , O= -2 , C= +4

because the C is positive atom it  will be the central atom with a double bond to the O and a single bond to each F.

this way all atoms completed the octet and stability

Answer:

Explanation:

If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane  a primary alkyl-halide.

Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used  otherwise it will result to a reactive mechanism if a weak electrophile is used.

However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.

The attached diagrams portraying this notions is shown in the attached file below.

Answer:

20.25 mL.

Explanation:

The volume of base required for the titration can be derived by removing the base titrant volume from the volume at endpoint.

i.e Final volume - initial volume

= (22.08 - 1.83)mL

= 20.25 mL.

(Repeat and average volume results for accuracy.)

Answer:

The pineal, hypothalamus, and pituitary are located in the BRAIN

The thyroid, parathyroid, and thymus are located in the THROAT

The adrenals, pancreas, testes, and ovaries are located in the NEAR THE KIDNEY

Explanation:

The pineal, hypothalamus, and pituitary are located in the BRAIN

The thyroid, parathyroid, and thymus are located in the THROAT

The adrenals, pancreas, testes, and ovaries are located in the NEAR THE KIDNEY

Answer:

Throat

Abdomen

Brain

Explanation:

These are correct!

Answer:

0.0289 kg of CO2 will be formed

Explanation:

Step 1: Data given

Mass of iron produced = 24.42 grams

Atomic mass iron = 55.845 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles iron

Moles iron = mass iron / molar mass iron

Moles iron = 24.42 grams / 55.845 g/mol

Moles iron = 0.437 moles

Step 4: Calculate moles CO2

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.437 moles Fe we'll have 3/2 * 0.437 = 0.6555 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.6555 moles * 44.01 g/mol

Mass CO2 = 28.85 grams = 0.0289 kg

0.0289 kg of CO2 will be formed

Answer:

See attached picture.

Explanation:

Hello,

In this case, given the compounds we notice:

a. Since carbon has four valence electrons (group IVA), it has three hydrogen atoms around it as well as a fluorine atom which has seven valence electrons (group VIIA). Moreover, we notice both carbon and fluorine attaining the octet.

b. In this case, two nitrogen atoms which have five valence electrons each (groupVA) are bonded via a single bond and two hydrogen atoms are bonded to each nitrogen atom in order to attain the octet for nitrogen only.

Thus, structures are given on the attached picture.

Regards.

Lewis structure for each covalent molecule CH₃F and N₂H₄ are attached to the image below.

a. CH₃F:

Carbon (C) has 4 valence electrons, Hydrogen (H) has 1 valence electron, and Fluorine (F) has 7 valence electrons.

The total number of valence electrons in CH3F can be calculated as follows:

1 carbon atom (C) × 4 valence electrons = 4 valence electrons

3 hydrogen atoms (H) × 1 valence electron = 3 valence electrons

1 fluorine atom (F) × 7 valence electrons = 7 valence electrons

Total valence electrons = 4 + 3 + 7 = 14

b. N₂H₄:

Nitrogen (N) has 5 valence electrons, and Hydrogen (H) has 1 valence electron.

The total number of valence electrons in N2H4 can be calculated as follows:

2 nitrogen atoms (N) × 5 valence electrons = 10 valence electrons

4 hydrogen atoms (H) × 1 valence electron = 4 valence electrons

Total valence electrons = 10 + 4 = 14

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Answer:

[tex]2.03~g~Ag_2S[/tex]

Explanation:

For this question we have to start with the reaction:

[tex]Ag~+~H_2S~+~O_2~->~Ag_2S~+~H_2O[/tex]

Now, we can balance the reaction, so:

[tex]4Ag~+~2H_2S~+~O_2~->~2Ag_2S~+~2H_2O[/tex]

With this in mind, we have to start with the amount of [tex]H_2S[/tex]. The first step is to convert from grams to moles. For this, we need to find the molar mass of [tex]H_2S[/tex]. If we check the periodic table we will find the atomic masses for Ag and H; H: 1 g/mol and A: 32 g/mol, so:

(1*2)+ (32*1) = 34 g/mol.

Now we can calculate the moles of [tex]H_2S[/tex]:

[tex]0.280~g~H_2S\frac{34~g~H_2S}{1~mol~H_2S}=0.0082~mol~H_2S[/tex]

With the moles of [tex]H_2S[/tex] we can calculate the moles of [tex]Ag_2S[/tex] if we check the molar ratio in the balanced equation, [tex]2~mol~H_2S=2~mol_Ag_2S[/tex], so:

[tex]0.0082~mol~H_2S\frac{2~mol_Ag_2S}{2~mol~H_2S}=0.0082~mol~H_2S[/tex]

With the molar mass of [tex]Ag_2S[/tex] we can convert from moles to grams (Ag: 107.86 g/mol, S: 32 g/mol), so:

(107.86*1)+(32*2)=247.80 g/mol

[tex]0.0082~mol~H_2S\frac{247.8~g~Ag_2S}{1~mol~H_2S}=2.03~g~Ag_2S[/tex]

I hope it helps!

Answer:

[Ni(CO)2C2O4]- Dicarbonyloxalatonickel(0)

K3[Fe(NCO)3(OCN)3-potassiumtricyanatotriisocyanatoferrate II

diammineaquahydroxoplatinum(II) ion- [Pt(NH3)2(H2O)(OH)]^+

Al[Co(CO3)2(CN)2- Aluminiumdicarbonatodicyanocobalt III

ethylenediaminedinitritozinc(II)- Zn(en)(NO2)2

Explanation:

The international Union of Pure and applied chemistry published a set rules for the nomenclature of inorganic compounds in 1985. This set of rules have been consistently revised ever since.

The rules stipulate the order for naming positive ions, negative ions and neutral ligands and the allowed order of preference for naming. It also stipulates the proper endings for ligands which are positive ions, negative ions or neutral ligands as the case may be.

The compounds named above were named in accordance with the revised IUPAC nomenclature for inorganic compounds.

Answer:

Explanation:Because of the delocalised electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules.

Answer:

Benzene has a pi electron system that is electron rich.

Electrophiles are molecules that are attracted to electron rich molecules as they are electron deficient in order to form a more stable complex.

Hence, as benzene is electron rich, electrophiles will be attracted to it.

Answer: b. -100 kJ

Explanation:

Heat of reaction is the energy released or absorbed during the course of the reaction. It is represented by the symbol [tex]\Delta H[/tex].

[tex]\Delta H=H_{products}-H_{reactants}[/tex]

[tex]\Delta H[/tex] = enthalpy of reaction = ?

[tex]H_{products}[/tex] = enthalpy of products = 0 kJ

[tex]H_{reactants}[/tex] = enthalpy of reactants = 100 kJ

[tex]\Delta H=0kJ-100kJ[/tex]

[tex]\Delta H=-100kJ[/tex]

Thus the ΔH for the reaction is -100kJ

The element fluorine is more electronegative than the element chlorine. Fluorine has an electronegativity value of 4.0 and Carbon has an electronegativity value of 2.5.

The term electronegativity is defined as a chemical property that describes the capacity of an atom or a functional group to attract electrons toward itself.

Electronegativity is refer to the ability of an atom to attract shared electrons in a covalent bond. The grater the value of the electronegativity, the more strongly that element attracts the shared electrons.

In fluorine electronegativity value is more than carbon. Therefore, it strongly attracts the shared electrons.

Thus, The element fluorine is more electronegative than the element chlorine.

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Answer:

The volume of the sample given is 850 ml, the density given is 0.79 gram per cm. Now the weight of the sample will be,  

Weight = volume × density = 850 × 0.79  

= 671.5 grams

Weight of the suspended solids given is 0.001 gram

The concentration of the sample can be determined by using the formula,  

Concentration = wt. of sample/volume

= [671.5 - 0.001) 10³ mg / 0.85 L

= 789998.82 mg/L or 789998.82 ppm

Now the concentration of suspended solids is.  

Css = 0.001 × 10³ mg / 0.85 L = 1.1764 mg per L or 1.1764 ppm

Answer:

6.24 x 10-3 M

Explanation:

Hello,

In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:

[tex]Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}[/tex]

Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change [tex]x[/tex], we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

[tex]5.2x10^{-5}=\frac{x*x}{0.749-x}[/tex]

Thus, we obtain a quadratic equation whose solution is:

[tex]x_1=-0.00627M\\x_2=0.00624M[/tex]

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.

Best regards.

Answer:

6.24×10¯³ M

Explanation:

Step 1:

The equation for the reaction is given below:

HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

Step 2:

The following data were obtained from the question:

Equilibrium constant, Kc = 5.2x10^-5

Concentration of HBrO, [HBrO] = 0.749 M

Concentration of BrO-, [BrO-] =?

Step 3:

Determination of the concentration of

BrO-.

We can obtain the concentration of BrO- as follow:

Initial concentration:

[HBrO] = 0.749 M

[H3O+] = 0

[BrO-] = 0

During reaction:

[HBrO] = - y

[H3O+] = +y

[BrO-] = +y

After reaction:

[HBrO] = 0.749 - y

[H3O+] = y

[BrO-] = y

Applying the equation constant equation:

Kc = [H3O+] [BrO-] / [HBrO]

5.2x10^-5 = y × y / 0.749

5.2x10^-5 = y² / 0.749

Cross multiply

y² = 5.2x10^-5 x 0.749

Take the square root of both side

y = √(5.2x10^-5 x 0.749)

y = 6.24×10¯³ M

[BrO-] = y = 6.24×10¯³ M

Therefore, the the concentration of BrO- at equilibrium is 6.24×10¯³ M

Answer:

4000cm³

Explanation:

2.0moles-1000cm³

8.0moles-?

(8.0×1000)÷2

=4000cm³

Explanation:

First off, it's important to know that the three subatomic particles we have are the proton, neutron and electron.

have a relative charge of +1 : This is the proton. It is the postive subatomic particle present in an atom.

have a relative charge of -1 : This is the electron. It is the negative subatomic particle present in an atom.

have no charge : This is the neutron. It is the neutral subatomic particle present in an atom.

located in the nucleus of an atom : Twoo subatomic particles are located in the nucleus, the neutron and proton.

have a much lower mass than the other two types of particles: This is the electron. The proton and neutron are almost similar in mass. The electron is abut 1 / 1840 times the mass of the proton.

Answer:

Number of Significant Figures: 5

The Significant Figures are 1 1 0 0 5

Explanation:

Answer:

I think first one has net charge 4+

second one has earlier 3+ charge after gaining net charge becomes zero

I think this is correct

Explanation:

if wrong means pls correct me bro

An atom has 22 protons and 19 electrons. The net charge on the atom is now elementary, 1e+. If the same atom with 22 electrons and 19 electrons were to gain 3 electrons, the net charge on the atom would be, negative, -3.

Electrons are the positively charged particles that revolve around the negatively charged nucleus. Electrons revolve around the nucleus, and protons and neutrons are also present.

When the electrons gain electron they acquire negative charge and when they donate electrons it acquires positive charge.

It gains 3 electrons that why -3 charge.

Thus, the correct elementary charge, negative, -3.

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Answer:

80 ml.

Explanation:

The balanced equation is:

KOH + HNO3 ---->  KNO3 + H2O

So 1 mole of KOH reacts with 1 mole of HNO3

200ml of 0.16M KOH contains 0.16 * 200/1000

= 0.16 * 0.2

= 0.032 M KOH.

So we need the volume (V) of 0.4 M HNO3 which contains 0.032 moles.

0.4 * V = 0.032

V = 0.032 / 0.4

V = 0.08 liters

= 80 mls.

Answer:

0.76 mole of Fe2S3.

Explanation:

Step 1:

Determination of the number of mole in 449g iron(III)bromide, FeBr3. This is illustrated below:

Mass of FeBr3 = 449g

Molar mass of FeBr3 = 56 + (80x3) = 296g/mol

Mole of FeBr3 =..?

Mole = Mass /Molar Mass

Mole of FeBr3 = 449/296

Mole of FeBr3 = 1.52 moles

Step 2:

The balanced equation for the reaction. This is given below:

2FeBr3 + 3Na2S —> 6NaBr + Fe2S3

Step 3:

Determination of the number of mole of Fe2S3 produced from the reaction of 449g ( i.e 1.52 moles) of FeBr3. This is illustrated below:

From the balanced equation above,

2 moles of FeBr3 reacted to produce 1 mole of Fe2S3.

Therefore, 1.52 moles of FeBr3 will react to produce = (1.52 x 1)/2 = 0.76 mole of Fe2S3.

Therefore, 0.76 mole of Fe2S3 is produced from the reaction.

Answer: 0.76

Explanation:

Answer:

P₂ = 3.24atm

Explanation:

Answer: 0.05 moles

Explanation:

To convert from atoms to moles, you will need Avogadro's number.

Avogadro's number: 6.022×10²³atoms/moles

(3.13×10²²atoms)×(moles/6.022×10²³ atoms)=0.05 moles

The atoms would cancel out and we will be left with our moles.

Answer:

3920kg of lime(CaO)   1568m^3 of CO2

Explanation:

I attached the explanation!

also for calculating the volume I assumed Standard Temperature and Pressure(STP).

Answer:

The correct answer is -0.129 kJ

Explanation:

In the given case, the cooling of the balloon is done by withdrawing 0.784 J of heat, and the work done by the atmosphere on the balloon is 655 J. First, there is a need to transform kJ into J, 1 kJ = 1000 J. So, 0.784 kJ would be 784 J.  

The ΔE or the change in the internal energy can be calculated by using the formula, ΔE = q + w ----- (1).  

In the given case, q refers to the heat moved out of the system, that is, the value of q would be less than 0 or will be a negative quantity. Therefore, the heat moved out of the system will be -784 J. On the other hand, as the work or w is done on the system, therefore, the value of w would be more than 0 or will be a positive quantity. Thus, the value of w will be +655 J.  

Now putting the values in the equation (1) we get,  

ΔE = -784 J + 655 J

ΔE = -129 J or -0.129 kJ

As the change in internal energy comes out to be a negative value, therefore, the process is considered exothermic.  

Answer:

I.  1.743mg/mL

II. 0.174g

III. 8.7 wt%

Explanation:

As absorbance is directly proportional to concentration. The calibration curve was:

Y = 1.6421X - 0.1113

Where Y is absorbance of sample and X its concentration in mg/mL

The diluted sample obtains an absorbance of 0.461, that means its concentration is:

0.461 = 1.6421X - 0.1113

0.3485mg/mL = X

I. The sample was diluted to 50.0mL from an aliquot of 10.0mL, thus, concentration in the original fruit sample was:

0.3485mg/mL × (50.0mL / 10.0mL) = 1.743mg/mL

II. The total mass of vitamin C in the sample (100.0mL) is:

100.0mL × (1.743mg/mL) = 174mg = 0.174g

III. As the fruit sample was of 2.0g, the wt% is:

0.174g / 2.0g = 8.7 wt%