An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically, but it can also be calculated using calculus. The reaction of compound A forming compound B was studied and the following data were collected:

Time (s) [A] (M)
0. 0.184
200. 0.129
500. 0.069
800. 0.031
1200. 0.019
1500. 0.016

Required:
a. What is the average reaction rate between 0 and 1500s?
b. What is the average reaction rate between 200. and 1200s ?
c. What is the instantaneous rate of the reaction at t=800s?

Answer:

Carbon and oxygen are chemically bonded in it.

Explanation:

The other answer choices do not apply for compounds, but rather for mixtures instead.

The  uncertainties  of  the  delta  measurements  and  the  uncertainty  of  the  atomic  weight  derivedfrom  the  best  measurement  of  isotopic  abundances  constrain  the  number  of  significant  figures  in  theatomic-weight values of the upper and lower bounds. For carbon, the fifth digit after the decimal pointis uncertain because of the uncertainty value of 0.000 027. Therefore, the number of significant digitsin the atomic-weight value is reduced to four figures after the decimal point. The Commission may rec-ommend  additional  conservatism  and  reduce  the  number  of  significant  figures  further.  For  the  lowerbound of carbon, 12.009 635 is truncated to 12.0096. For an upper bound, the trailing digit is increasedto ensure the atomic-weight interval encompasses the atomic-weight values of all normal materials. Inthe case of carbon, the upper bound is adjusted from 12.011 532 to 12.0116 to express four digits afterthe decimal point. The lower and upper bounds are evaluated so that the number of significant digits ineach  is  identical.  If  a  value  ends  with  a  zero,  it  may  need  to  be  included  in  the  value  to  express  therequired number of digits. The following are examples of lower and upper atomic-weight bounds foroxygen that could be published by the Commission in its various tables.

---------------------------

Extracted from" Atomic weights of the elements 2009 (IUPAC Technical Report)"

Answer:

The soil has a percentage of water by mass of 27.8 %.

Explanation:

Keeping in mind that

we can calculate the mass of water:

We can then calculate the mass percentage of water in the soil:

Answer:

Molecules/State

Explanation:

We know from the kinetic molecular theory that matter is made up of molecules. These molecules are in constant motion at various velocities and energies.

The total thermal energy refers to the total kinetic energies of these particles. It depends on the temperature of the system, the energies of the molecules present in the material and the arrangement of the object's molecules (states of matter).

Answer: product

Explanation:

Each substance written to the right of the arrow in a chemical equation is referred to as a product.

When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.

On the other hand which is the right side is the product.

Answer:

V₂ = 104.76 mL

Explanation:

Given data:

Initial volume = 100.0 mL

Initial temperature = 21°C (21 + 273.15 K = 294.15 K)

Final temperature = 35°C (35 + 273.15 K = 308.15 k)

Final volume = ?

Solution:

Charles Law:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ =100.0 mL × 308.15 K / 294.15 K

V₂ = 30815 mL.K /294.15 K

V₂ = 104.76 mL

Answer:

1.2 cm

Explanation:

The reason how I got my answer is by dividing 150 by 125 and that gave me 1.2 cm.

Answer:

Liquid

Explanation:

Answer:

See explanation and image attached

Explanation:

Often times, there is a need for a three dimensional representation of a molecule on paper. These three dimensional representations give us an idea of what the molecule really looks like if we were to be looking at it physically.

In order to make a three dimensional representation, we use wedged and dashed bonds. The wedged bonds are coming out of the plane of the paper towards you while the dashed bonds are going into the plane of the paper away from you.

In the image attached, you will find the three dimensional representation of the methane molecule.

Answer:

th relative ability of a solute to devolve into a solvent

Answer:

The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.

Explanation:

Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.

Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).

These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.

Therefore,  Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

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Answer:

116.78 grams.

Explanation:

1 mol of Strontium (Sr) = 87.62 grams

1.38 mol of Strontium = x

Cross Multiply

1 * x = 1.38 * 87.62

x = 116.78  grams

Answer:

1.17 M

Explanation:

Step 1: Write the balanced equation

N₂ + 3 H₂ ⇒ 2 NH₃

Step 2: Calculate the rate of disappearance of H₂

We will use the following expression.

rH₂ = - Δ[H₂]/t = - (0.25 M - 0.50 M)/44 s = 0.0057 M/s

Step 3: Calculate the rate of disappearance of N₂

The molar ratio of N₂ to H₂ is 1:3.

0.0057 mol H₂/L.s × 1 mol N₂/3 mol H₂ = 0.0019 mol N₂/L.s

Step 4: Calculate the final concentration of N₂

We will use the following expression.

[N₂] = [N₂]₀ - rN₂ × t

[N₂] = 1.25 mol/L - 0.0019 mol/L.s × 44 s

[N₂] = 1.17 M

Answer:

An ionic bond is the electrostatic force that holds ions together when they form bonds

Explanation:

An ionic bond is formed when a metal looses electron(s) to a non metal leading to the formation of a positive ion and a negative ion.

An ionic compound is actually an ion pair, the ions are held together by strong electrostatic forces.

This strong electrostatic force that holds the ion pair together in ionic compounds is what we commonly call the IONIC BOND.  

Answer:

An ionic bond is the electrostatic force that holds ions together when they form bonds

Explanation:

I took this test and it was the correct answer :)

Answer:

Observational evidence is essential for investigating the way disease affects populations, the patterns and distribution of risk within them, and the emergence of trends in health and disease over time.

Answer:

It tests a prediction It supports the results. It asks a testable question It predicts what will happen

Explanation:

Answer:

[N₂] = 1.1M

Explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

Where [] are equilibrium concentrations of each specie

As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M

Answer:

Acidic solution used to clean a crucible

Explanation:

This liquid dissolves alcoholic solvents such as crucible, that is why it was selected as the ideal for cleaning.

Although it would be ideal to know in detail which chemical compound is the one you want to clean so that the cleaning technique has better effectiveness.

WASTE CONTAINER refers to the solvent used to rinse chemicals out of a beaker, and it also refers to the acid solution used to clean a crucible. Water used to rinse the detergent out of a flask refer to the SINK.

In conclusion, WASTE CONTAINER refers to the solvent used to rinse chemicals out of a beaker, and it also refers to the acid solution used to clean a crucible. Water used to rinse the detergent out of a flask refer to the SINK.

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Answer:

An ecosystem is a community or group of living organisms that live in and interact with each other in a specific environment.

Explanation:

That's the most straight forward explanation I could come up with

The percent composition of the compound.

A O-18.18%, N-21.21%, H-60.60%

Given

6.00 grams of oxygen,

7.00 grams of nitrogen,

20.00 grams of hydrogen.

Required

The percent composition

Solution

Total mass :

= mass of O + mass of N + mass of H

= 6 + 7 + 20

= 33 g

% O = 6/33 x 100%= 18.18%

% N = 7/33 x 100%=21.21%

% H = 20/33 x 100% = 60.6 %

Answer:

H-H equation is written as follows:

pH=pK + log

{HCO3-}(base)

{H2CO3}(acid)

Answer:

5.78 × 10¹⁴ Hz

Explanation:

Step 1: Given and required data

Step 2: Convert "λ" to meters

We will use the conversion factor 1 m = 10⁹ nm.

519. nm × 1 m/10⁹ nm = 5.19 × 10⁻⁷ m

Step 3: Calculate the frecquency of this light

We will use the following expression.

c = λ × ν

ν = c/λ

ν = (3.00 × 10⁸ m/s)/5.19 × 10⁻⁷ m

ν = 5.78 × 10¹⁴ s⁻¹ = 5.78 × 10¹⁴ Hz

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

Answer:

b

Explanation:

Answer:

0.229 cm³.

Explanation:

The following data were obtained from the question:

Volume (in in³) = 0.014 in³

Volume (in cm³) =?

1 in = 2.54 cm

Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:

1 in = 2.54 cm

Therefore,

1 in³ = 2.54³ cm³

1 in³ = 16.387 cm³

Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:

1 in³ = 16.387 cm³

Therefore,

0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³

0.014 in³ = 0.229 cm³

Thus, 0.014 in³ is equivalent to 0.229 cm³.

Answer:

Q =  57934.8 J

Explanation:

Given data:

Mass of water = 180.0 g

Initial temperature = 19°C

Fina temperature = 96°C

Specific heat capacity of water = 4.18 J/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 96°C - 19°C

ΔT = 77°C

Q = 180.0 g×4.18 J/g.°C×77°C

Q =  57934.8 J

Answer:

Explanation:

Nitric oxide is the gas NO, it reacts with oxygen as shown below;

2NO(g) + O2(g) -----> 2NO2(g)

Now the gas formed is the gas NO2 which is known to be reddish brown in colour.

A diagrammatic representation of this reaction is shown in the image attached to this answer.

Image credit: Chemlibretext

Answer:

d. 9.95 × 10⁻³ mol

e. 249 g

f. 2.55 × 10²⁴ atoms

g. 4.32 × 10²² atoms

Explanation:

d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol

e. the mass of 4.23 mol of cobalt

The molar mass of cobalt is 58.93 g/mol.

4.23 mol × 58.93 g/mol = 249 g

f. the number of cobalt atoms in 4.23 mol of cobalt

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms

g. the number of cobalt atoms in 4.23 g of cobalt

First, we will calculate the moles of cobalt using the molar mass of cobalt.

4.23 g × 1 mol/58.93 g = 0.0718 mol

Then, we will calculate the number of cobalt atoms using Avogadro's number.

0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms

Answer:

The answer is "A"

Explanation:

Please find the complete question in the attachment file.

[tex]\to R(s)= \frac{As}{K+s}[/tex]

when the s in the approach, that is infinity R(s) tends

[tex]\to \frac{A}{\frac{K}{s}+1} \\\\ \to\frac{A}{0+1} \\\\ \to\frac{A}{1} \\\\ \to A[/tex]

Answer:

Solution A: 0.00400M

Solution B: 0.00400M

Solution C: 4.00x10⁻⁵M

Explanation:

Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:

250mL / 10mL = 25 times.

That means molar concentration of sln A is:

0.100M / 25 = 0.00400M

Solution B is obtained diluting 25mL to 100mL:

100mL / 25mL = 4 times

0.00400M / 4 times = 0.00100M

And solution C is obtained diluting the solution C from 20mL to 500mL:

500mL / 20mL = 25 times

Solution C:

0.00100M / 25 times = 4.00x10⁻⁵M

The formula for serial dilution can be used to obtain the molarity of solution A, B , C.

M1V1 = M2V2

M2 = 0.100 M ×  10 mL/250-mL

M2 = 0.004 M

M1V1 = M2V2

M2 = 0.004 M × 25 mL/100-mL

M2 = 0.001 M

M1V1 = M2V2

M2 = 0.001 M × 20 mL/500-mL

M2 = 0.00004 M

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