Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable.
(i) 0.200 m HOCH2CH2OH
(ii) 0.200 m Ba(NO3)2
(iii) 0.200 m K3PO3
(iv) 0.200 m Ca(CIO4)2
(v) These all have the same freezing point.

Answer:

298. 7 K.

Explanation:

Hello!

In this case, since equation we use to compute the heat in a cooling or heating process is:

[tex]Q=mC(T_f-T_i)[/tex]

Whereas we are given the heat, mass, specific heat and initial temperature. Thus, we infer that we need to solve for the final temperature just as shown below:

[tex]T_f=T_i+\frac{Q}{mC}\\\\T_f=1000 K+\frac{-270000J}{1000g*0.385\frac{J}{g*K} } \\\\T_f=298.7 K[/tex]

It is important to notice that the iron release heat as water absorbs it, that is why it is taken negative.

Best regards!

Answer:

The differential equation describing the given situation is dQ/dt = k( 16 - Q )

Explanation:

Given the data in the question;

Initially, the water based solution contains 16 grams of undissolved chemicals;

Assume Q(t) is the amount of dissolved chemical at time L

then the amount of undissolved chemicals at time t is ( 16 - Q)

The rate of change of amount of chemicals dissolved in the solution is proportional to the amount of undissolved chemicals

this means;

dQ/dt ∝ ( 16 - Q)

dQ/dt = k( 16 - Q )

where k is the positive proportionality constant.

Therefore, The differential equation describing the given situation is dQ/dt = k( 16 - Q )

Answer:

Formulas

3.2 Determining Empirical and Molecular Formulas

Learning Objectives

By the end of this section, you will be able to:

Compute the percent composition of a compound

Determine the empirical formula of a compound

Determine the molecular formula of a compound

The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.

Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

%H=mass Hmass compound×100%

%C=mass Cmass compound×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

%H=2.5g H10.0g compound×100%=25%

%C=7.5g C10.0g compound×100%=75%

EXAMPLE 3.9

Calculation of Percent Composition

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

Solution

To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

%C=7.34g C12.04g compound×100%=61.0%%H=1.85g H12.04g compound×100%=15.4%%N=2.85g N12.04g compound×100%=23.7%

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

Check Your Learning

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

ANSWER:

12.1% C, 16.1% O, 71.8% Cl

Determining Percent Composition from Molecular or Empirical Formulas

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

%N=14.01amu N17.03amuNH3×100%=82.27%%H=3.024amu H17.03amuNH3×100%=17.76%

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.10. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the

[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]

Reaction

2N2O(g) — 2N2(g) + O2(g)

Required

relative rate

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

so the relative rates for the reaction above are :

[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]

Answer:

A. Ba

Explanation:

Atomic radius increases as you go down and decreases as you go right.

The empirical and  molecular formula : C₂₁H₂₂N₂O₂

Given

The percent composition of an unknown substance is 75.42% Carbon, 6.63 % Hydrogen, 8.38 % Nitrogen, and 9.57 % Oxygen

Required

The empirical and  molecular formula

Solution

C : 75.42 : 12 = 6.285

H : 6.63 : 1 = 6.63

N : 8.38 : 14 = 0.599

O : 9.57 : 16 = 0.598

Divide by 0.598

C : H : N : O = 10.5 : 11 : 1 : 1 = 21 : 22 : 2 : 2

The empirical formula : C₂₁H₂₂N₂O₂

(C₂₁H₂₂N₂O₂)n = 334

(334)n=334

n = 1

Answer:

Same bond angle-CO2, CH4, SF6, PCl5, BF3, BeF2, XeF4,XeF2

Different bond angle- H2O, SO2, NH3, ClF3,SF4

Explanation:

We know that the shape of molecules and and bond angles between atoms in molecules are predicted on the basis of the valence shell electron pair repulsion theory.

Most times, certain molecules deviate from the expected bond angles for different reasons. The most common reason is the presence of lone pairs on the molecule. This is the case in the molecules; H2O, SO2, NH3, ClF3 and SF4.

However, in XeF4 and XeF2, the lone pairs orient themselves in such a way that they do not distort the expected bond angle. For instance, in XeF2, the  three lone pairs occupy equatorial positions while the two bond pairs occupy apical positions. In XeF4, the bond pairs are directed at the corners of a square while the two lone pairs are positioned above and below the plane of the square.

Answer:

4.82 moles of Ag.

Explanation:

We'll begin by calculating the number of mole in 450.98 g of Cu(NO₃)₂. This can be obtained as follow:

Molar mass of Cu(NO₃)₂ = 63.5 + 2[14 + (16×3)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g/mol

Mass of Cu(NO₃)₂ = 450.98 g

Mole of Cu(NO₃)₂ =?

Mole = mass /Molar mass

Mole of Cu(NO₃)₂ = 450.98 / 187.5

Mole of Cu(NO₃)₂ = 2.41 moles

Next, we shall determine the number of mole of Cu needed to produce 450.98 g (i.e 2.41 moles) of Cu(NO₃)₂. This can be obtained as follow:

Cu + 2AgNO₃ —> Cu(NO₃)₂ + 2Ag

From the balanced equation above,

1 mole of Cu reacted to produce 1 mole of Cu(NO₃)₂.

Therefore, 2.41 moles of Cu will also react to produce 2.41 moles of Cu(NO₃)₂.

Thus, 2.41 moles of Cu is needed for the reaction.

Finally, we shall determine the number of mole of Ag produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of Ag.

Therefore, 2.41 moles of Cu will react to produce = 2× 2.41 = 4.82 moles of Ag.

Thus, 4.82 moles of Ag were obtained from the reaction.

Answer:

it develops from the womb

Answer:

outer cells of the blastocyst

Explanation:

on edg:)

Answer:

b. temperature difference

Answer:

what is that's a question

According to specific heat capacity, 13.09 grams of water at 0 degree Celsius  can be frozen into ice at zero degree c if 55 kJ of heat is removed.

Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.

It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.

It is given by the formula ,

Q=mcΔT, substitution in formula gives Q/c=m which is m= 55/4.2=13.09 grams.

Thus, 13.09 grams of water at 0 degree Celsius  can be frozen into ice at zero degree c if 55 kJ of heat is removed.

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458 g H₂SO₄

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

4.67 mol H₂SO₄

Step 2: Identify Conversions

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of S - 32.07 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂SO₄ - 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

458.08 g H₂SO₄ ≈ 458 g H₂SO₄

Answer:

3 H1 NMR signals

Explanation:

NB: kindly check the diagram of the chemical compound in the attached picture.

This particular Question is based on the part of chemistry which is known as spectroscopy. Spectroscopy is used in the Determination or in identifying chemical compounds. H'NMR works on the principle of nuclear magnetic resonance.

In order to solve this question, one has to count the number of hydrogen in unique location. The diagram in the attached show how hydrogen is been counted.

The numbers of signals is the number of different chemical environments in which hydrogen atoms are located.

NB: signals is also the same as peak in H'NMR.

Hence, the number of H1 NMR signals in this chemical compound is 3.

Answer:

Enzyme mutations can lead to serious or fatal human disorders and are the consequence of inherited abnormalities in the affected individual's DNA. The mutation may be at a specific position in an enzyme encoded by a mutated gene, just like a single abnormal amino acid residue.

Explanation:

Answer:

Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy.

...

Energy Levels of Electrons.

Energy Level Energy

1 -13.6 eV

2 -3.4 eV

3 -1.51 eV

4 -.85 eV

Answer: i think its A diatomic compound..

Explanation: hope i helped! sorry if im wrong!

Answer:

C8H16O4

Explanation:

C2H4O= 24+4+16

44

n=molar mass/empirical formula

n=176.21/44

=4

Therefore

Molar Formula= (C2H4O)4=C8H16O4

Answer:

Explanation:

Energy of falling radiation having wavelength of 265 nm

= h c / λ where h is plank's constant , c is velocity of light and λ is wavelength of radiation . Putting the values

Energy of light photon = 6.6 x 10⁻³⁴ x 3 x 10⁸ / 265 x 10⁻⁹

= .0747 x 10⁻¹⁷

= 7.47 x 10⁻¹⁹ J .

Work function of sodium is 4.41 x 10⁻¹⁹

So kinetic energy of ejected electron = energy of falling photon - work function

= 7.47 x 10⁻¹⁹ - 4.41 x 10⁻¹⁹

= 3.06 x 10⁻¹⁹ eV .

Answer:

Dispersion forces

Dipole-Dipole interaction

Explanation:

The London dispersion force refers to the temporary attractive force that acts between the electrons in two adjacent atoms when the atoms develop temporary dipoles. Dispersion forces act between any two molecules even when other intermolecular forces are in operation as long as the molecules are in close proximity to each other.

Now, CO is polar and the HCN is also polar molecule. Hence, dipole - dipole interaction forces  are also in operation and acts between the two molecules in close proximity to each other.

Dispersion forces and Dipole-Dipole interaction are intermolecular forces which act between a hydrogen cyanide (HCN) molecule and a carbon monoxide molecule

The transitory attractive force that exists between the electrons in two nearby atoms when the atoms form transient dipoles is known as the London dispersion force. As long as the molecules are close to one another, dispersion forces can exist between any two molecules, even when other intermolecular forces are active.

The HCN molecule and CO are both polar molecules right now. As a result, dipole-dipole interaction forces act between the two molecules when they are close to one another.

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Answer:

c. bigger planets have more gravity and smaller planets have less gravity. i hope this helped :))

Answer:

i hope my answer help u :))

Explanation:

no , it will be 39.99711

Answer:

BrO3

Explanation:

The empirical formula is the smallest whole-number ratio, you find the greatest common factor (which is 2, in this case), then divide the subscripts by it.

So:

Br2 / 2 = Br1

O6 / 2 = O3

Answer:

1.01 grams/ cm^3

Explanation:

because that's just how it is

Answer:

An investigation is about collecting information, while an experiment is testing  something based off an investigation

Explanation:

Hope this helps

An experiment tests something based on an investigation, whereas an investigation pertains to gathering knowledge.

The operation of something is demonstrated. There are independent and dependent variables in an experiment. The control of variables distinguishes an experiment from a demonstration. Add an independent as well as a dependent variable to the project in order to convert a demonstration into an experiment.

An information resource's value to a particular professional group and its influence on the procedures and results of healthcare are just a few of the problems that demonstration studies address. An experiment tests something based on an investigation, whereas an investigation pertains to gathering knowledge.

Therefore, an experiment tests something based on an investigation, whereas an investigation pertains to gathering knowledge.

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Answer:

Motion control systems are any system that control one or more of the following of a machine: its position, velocity, force, and/or pressure. They generally consist of the follow components: Motion controller: the central part that operates the system (I.e. the brain)

Here, we are required to determine what types of atoms are present in the simplest chemical substance.

Chemical substances are generally classified as either;

Definition of terms;

Elements and Compounds are collectively termed Pure Substances.

Therefore, inference drawn from the premises above point to the fact that the simplest chemical substances are composed of the atoms of one type of element.

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