A simplified version of photosynthesis can be represented as carbon dioxide
combining with water to form glucose and oxygen: 6CO2 + 6H20 - C6H12O6 +
602. In this reaction, is oxidized.
A) hydrogen
B)
oxygen
C) carbon dioxide
D) carbon

[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:

it develops from the womb

Answer:

outer cells of the blastocyst

Explanation:

on edg:)

Answer: i think its A diatomic compound..

Explanation: hope i helped! sorry if im wrong!

Answer:

i hope my answer help u :))

Explanation:

no , it will be 39.99711

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:

A. Ba

Explanation:

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

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 )

[tex]{ \boxed{ \bold{ \blue{Question}}}}[/tex]

Is H2SO4 an element, compound or molecule?

[tex]{ \boxed{ \bold{ \blue{Answer}}}}[/tex]

H2SO4 (sulphuric acid) is a compound. It is made up of 2 hydrogen atoms, 1 sulphur atom, and 4 oxygen atoms.

_______________________________

[tex] \underline \bold \orange{hope \: it \: helps}[/tex]

A chemical compound is any substance that consists of two or more different elements combined together.

I think that H2SO4 fits this definition of a compound .

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️

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:

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

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:

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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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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Answer:

[tex]m_{leftover}=37g[/tex]

Explanation:

Hello!

In this case, since the melting process involves the heat added to the system, the heat of fusion and the mass of water:

[tex]Q=m\Delta H_{fus}[/tex]

Thus, as it has been reported that the heat of fusion of ice is 333.6 J/g, we can compute the melted mass of ice as shown below:

[tex]m=\frac{Q}{\Delta H_{fus}} =\frac{46000J}{333.6J/g}\\\\m=138g[/tex]

Thus, the ice leftover results:

[tex]m_{leftover}=175g-138g\\\\m_{leftover}=37g[/tex]

Regards!

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:

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:

12408 feet per minute

Explanation:

Given: Speed is 141 mi/h

To find: speed in units of feet per minute

Solution:

Use the following units to convert the given speed into feet per minute.

1 mile = 5280 foot

1 h = 60 minutes

Therefore,

141 mi/h [tex]=\frac{141(5280)}{60} =12408[/tex] feet per minute.

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:

1.01 grams/ cm^3

Explanation:

because that's just how it is

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:

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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Answer:

SnCl4 * 5H2O

Explanation:

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

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:

The answer is "Strontium iodide and [tex]SrI_2[/tex]".

Explanation:

In the flames test, two ionic metals give red fire. It was calcium (Ca, which gives a red brick flame) and strontium (Sr, which gives a persistent red flame). Both Ca and Sr respond to insoluble carbonate with ammonium carbonate. Ca and Sr produce insoluble ammonium phosphate phosphates. Even then, the water is reactive with calcium sulfate (CaSO4) whereas the strontium sulfate (SrSO4) is illiquid. Because the metal ion formed 3 precipitates, they can't get Ca, and we have Sr as its metal.

The hexane of its halide gives a violet color which would be typical of iodide (I-).

Strontium iodide is the chemical name of the unknown ionic compound.

Strontium iodide's chemical formula is [tex]SrI_2[/tex].