NEED HELP!
C=46.67%, H=4.48%, N=31.10%, O=17.76%.
The molecular weight is 180.16g/mol.

Answers

Answer 1

Answer:

Answer:

$C_7H_8N_4O_2$

Explanation:

Hello!

In this case, since the determination of an empirical formula is covered by first computing the moles of each atom as shown below:

$n_C=(46.47g)/(12g/mol)=3.9mol\n\n n_H=(4.48g)/(1g/mol) =4.5mol\n\nn_N=(31.10g)/(14g/mol) =2.2mol\n\nn_O=(17.76g)/(16g/mol) =1.1mol$

Now, we divide each moles by the fewest moles (those of oxygen), to obtain the subscripts in the empirical formula:

$C:(3.9)/(1.1)=3.5 \n\nH:(4.5)/(1.1)=4 \n\nN:(2.2)/(1.1) =2\n\nO:(1.1)/(1.1) =1$

Thus, the empirical formula, taken to the nearest whole subscript is:

$C_7H_8N_4O_2$

Whose molar mass is 180.16, therefore the empirical formula is the same to the molecular one.

Best regards!

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How many lone pairs are on the central atom of BrF3?

Answers

According to the molecular geometry, there are two lone pairs on central atom of BrF₃.

What is molecular geometry?

Molecular geometry can be defined as a three -dimensional arrangement of atoms which constitute the molecule.It includes parameters like bond length,bond angle and torsional angles.

It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.

They provide information about geometry by taking into considerations the vibrational and rotational absorbance of a substance.Neutron and electron diffraction techniques provide information about the distance between nuclei and electron density.

Learn more about molecular geometry,here:

brainly.com/question/28557524

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Answer:

BrF3 has 2 lone pairs (4 unshared electrons)

Explanation:

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Answers

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer:

$m_(PbI_2)=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_(Pb(NO_3)_2)=(0.14gPb(NO_3)_2)/(1g\ sln)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2) *(1.134g\ sln)/(1mL\ sln) *96.7mL\ sln\n\nn_(Pb(NO_3)_2)=0.04635molPb(NO_3)_2\n\nn_(KI)=(0.12gKI)/(1g\ sln)*(1molKI)/(166.0gKI) *(1.093g\ sln)/(1mL\ sln) *99.8mL\ sln\n\nn_(KI)=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_(PbI_2)=0.07885molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \n\nm_(PbI_2)=18.2gPbI_2$

Best regards.

Identify the true statement(s) about the valence bond theory. 1. The strength of a bond depends on the amount of overlap between the two orbitals of the bonding atoms. 2. The greater the overlap between two bonding atoms, the lesser the bond strength. 3. Orbitals bond in the directions in which they protrude or point to obtain maximum overlap

Answers

Answer:

The strength of a bond depends on the amount of overlap between the two orbitals of the bonding atoms

Orbitals bond in the directions in which they protrude or point to obtain maximum overlap

Explanation:

The valence bond theory was proposed by Linus Pauling. Compounds are firmed by overlap of atomic orbitals to attain a favourable overlap integral. The better the overlap integral (extent of overlap) the better or stringer the covalent bond.

Orbitals overlap in directions which ensure a maximum overlap of atomic orbitals in the covalent bond.

Answer:

THE STRENGTH OF THE BOND DEPENDS ON THE AMOUNT OF OVERLAP BETWEEN THE TWO ORBITALS OF THE BONDING ATOMS

ORBITALS BOND IN THE DIRECTION OR POINT IN WHICH THEY PROTRUDE OR POINT TO OBTAIN MAXIMUM OVERLAP.

Explanation:

Valence bond theory describes the covalent bond as the overlap of half-filled atomic orbital yields a pair of electrons shared between the two bonded atoms. Overlapping of orbitals occurs when a portion of one orbital and the other occur in the same region of space. The strength of a bond is determined by the amount of overlap between the two orbitals of the bonding atoms. In other words, orbitals that overlap more and in the right orientation of maximum overlapping form stronger bonds that those with less overlap and right orientation for maximum overlap. The bonding occurs at a varying distance in different atoms from which it obtains its stable energy caused by the increase in the attraction of nuclei for the electrons.

Orbitals also bond in the direction to obtain maximum overlap as orientation of the atoms also affect overlap. The greater overlap occurs when atoms are oriented on a direct line mostly end to end or side by side between the two nuclei depending on the type of bond formed. A sigma bond is formed when atoms overlap end to end in which a straight line exists between the two atoms that is the internuclear axis indicating the concentrated energy density in that region. Pi bond exits in when overlap occurs in the side -to -side orientation and the energy density is concentrated opposite the internuclear axis.

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.

Answers

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of $HNO_3$

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of $C_6H_5NH_2$ .

$M_2\text{ and }V_2$ are the final molarity and volume of $HNO_3$ .

We are given:

$M_1=0.3403M\nV_1=160.0mL\nM_2=0.0501M\nV_2=?$

Putting values in above equation, we get:

$0.3403M* 160.0mL=0.0501M* V_2\n\nV_2=1086.79mL$

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of $C_6H_5NH_2$ + Volume of $HNO_3$

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

$\text{Concentration of salt}=(0.3403M)/(1246.79mL)* 160.0mL=0.0437M$

Now we have to calculate the pH of the solution.

At equivalence point,

$pOH=(1)/(2)[pK_w+pK_b+\log C]$

$pOH=(1)/(2)[14+4.87+\log (0.0437)]$

$pOH=8.76$

$pH+pOH=14\n\npH=14-pOH\n\npH=14-8.76\n\npH=5.24$

Thus, the pH of the solution is, 5.24

What is the freezing point of a solution of 7.15 g MgCl2 in 100 g of water? K f for water is 1.86°C/m. What is the freezing point of a solution of 7.15 g MgCl2 in 100 g of water? K f for water is 1.86°C/m. -0.140°C -2.80°C -1.40°C -4.18°C

Answers

Answer:

THE NEW FREEZING POINT IS -4.196 °C

Explanation:

ΔTf = 1 Kf m

molarity of MgCl2:

Molar mass = (24 + 35.5 *2) g/mol

molar mass = 95 g/mol

7.15 g of MgCl2 in 100 g of water

7.15 g = 100 g

(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3

= 0.715 g /dm3

Molarity in mol/dm3 = molarity in g/dm3 / molar mass

= 0.715 g /dm3 / 95 g/mol

m = 0.00752 mol/ dm3

So therefore:

ΔTf = i Kf m

1 = 3 (1 Mg and 2 Cl)

Kf = 1.86 °C/m

M = 0.752 moles

So we have:

ΔTf = 3 * 1.86 * 0.752

ΔTf = 4.196 °C

The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C

What is the correct formula for phosphorus pentachloride? (a) PCl5 (b) P5Cl (c) P(ClO)5 (d) PO4Cl (e) PCIO

Answers

Answer:

(a) $PCl_5$

Explanation:

Hello,

In this case, such nomenclature expresses the amount of both phosphorous and chlorine atoms via prefixes for the sub indexes of each atom into the molecule, thus, as phosphorous is prefixless one infers that there is just one phosphorous and five chlorides since the prefix for such atom is penta. Therefore, the correct formula is:

$PCl_5$

Hence, the answer is (a)

Best regards.

Answer:

The correct formula for phosphorus pentachloride is a) PCl5

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