The half-life for a first-order reaction is 32 s. What was the original concentration if, after 2.0 minutes, the reactant concentration is 0.062 M?​

Answer:d8

Explanation:

Answer:

i have no clue

Explanation:

i have no clue

Explanation:

The reaction expression is given as:

           2H₂  +   O₂    →    2H₂O

The limiting reactant is the reactant in short short supply in the given reaction.

To find the limiting reactant we use the number of moles.

    Mass of H₂  = 5g

    Mass of O₂   = 10g

  Number of moles  = [tex]\frac{mass}{molar mass}[/tex]  

       Molar mass of H₂   = 2(1) = 2g/mol

       Molar mass of O₂  = 2(16)  = 32g/mol

Number of moles  H₂ = [tex]\frac{2}{5}[/tex]   = 0.4mole

Number of moles O₂ = [tex]\frac{10}{32}[/tex]   = 0.3mole

   From the balanced reaction equation:

           2 mole of H₂ reacts with 1 mole of O₂

          0.4 mole of H₂ reacts with  0.2 mole of O₂

But we are given 0.3 moles of O₂, so the limiting reactant is H₂ because oxygen gas is in excess

The limiting reactant when 5.00g of H₂ and 10.00g of O₂ react is Oxygen (O₂)

From the question,

We are to determine the limiting reactant

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

2H₂ + O₂ → 2H₂O

This means 2 moles of H₂ reacts with 1 mole of O₂ to produce 2 moles of H₂O

Now, we will determine the number of moles of each reactant present

Mass = 5.00 g

Molar mass = 2.0 g/mol

Using the formula,

[tex]Number\ of \ moles = \frac{Mass}{Molar\ mass}[/tex]

Number of moles of H₂ = [tex]\frac{5.00}{2.0}[/tex]

Number of moles of H₂ = 2.50 moles

Mass = 10.00 g

Molar mass = 32.0 g/mol

Then,

Number of moles of O₂ = [tex]\frac{10.00}{32.0}[/tex]

Number of moles of O₂ = 0.3125 mole

Now,
Since 2 moles of H₂ reacts with 1 mole of O₂

Then,

0.625 moles of H₂ reacts with 0.3125 mole of O₂

This means part of the H₂ reacts with all of the O₂. Since all the O₂ reacted, then O₂ is the limiting reactant.

Hence, the limiting reactant when 5.00g of H₂ and 10.00g of O₂ react is Oxygen (O₂).

Learn more on Stoichiometry here: https://brainly.com/question/17199947

Answer

false

Explanation:

sorry I tbought i put false I hope I helped ....

The answer is false! All matters do not look the same, even if it had a visible look the answer would be false!

Why is the answer false?

There is lots of matters, Like liquids:

(Water, Juice, Milk, etc.)

All Water, Juice, & Milk looks different, Maybe SOME look the same, but the question is saying "All", Most of all questions is wrong.

Therefore the answer is false!

Learn more about matter on www.Bing/LearnMore/Subject="Matter%20Learning"/Search.com

Need more of my help? Put #AJQ in your questions and I will come fast!

Answer:

64 k

Explanation:

To figure this out you need to start with the ideal gas equation:

P

⋅

V

=

n

⋅

R

⋅

T

You have P (3.30 atm), V (3.5 l), n (2.20 moles) and you can look up the gas constant, R (0.082057 (latm)/(molK)).

We simply rearrange the ideal gas equation to get T by itself:

T

=

P

⋅

V

n

⋅

R

Then simply plug in the values you were given, making sure that the units are in liters, atmospheres, and moles (which they are, in this case):

T

=

3.30

atm

⋅

3.5

l

2.20

moles

⋅

0.082057

(l*atm)/(mol*K)

=

63.9799

K

Since volume was only given to 2 significant figures, we can only report 2 significant figures for our answer:

64 K.

Answer:

[tex]\frac{1}{-1} \frac{d[A]}{dt} =\frac{1}{-3} \frac{d[B]}{dt}=\frac{1}{1} \frac{d[C]}{dt}=\frac{1}{2} \frac{d[D]}{dt}[/tex]

[tex]\frac{d[A]}{dt} =-0.9\frac{mol}{dm^3*s} \\\\ \frac{d[C]}{dt} =0.9\frac{mol}{dm^3*s}\\\\ \frac{d[D]}{dt} =1.8\frac{mol}{dm^3*s}[/tex]

Explanation:

Hello!

In this case, since the reaction rate for all the species are written by considering how they change and the coefficients they have in the reaction, for A, B, C and D we can write:

[tex]\frac{1}{-1} \frac{d[A]}{dt} =\frac{1}{-3} \frac{d[B]}{dt}=\frac{1}{1} \frac{d[C]}{dt}=\frac{1}{2} \frac{d[D]}{dt}[/tex]

Whereas the numeric denominators stand for the coefficients balancing the reaction. In such a way, by using the rate of consumption of B, we can compute the rate of consumption of A and the rates of formation of both C and D as follows:

[tex]\frac{d[A]}{dt} =\frac{-1}{-3} \frac{d[B]}{dt}=\frac{1}{3}*-2.7\frac{mol}{dm^3*s} \\\\ \frac{d[A]}{dt} =-0.9\frac{mol}{dm^3*s} \\\\\\ \frac{d[C]}{dt} =\frac{1}{-3} \frac{d[B]}{dt}=\frac{1}{-3}*-2.7\frac{mol}{dm^3*s} \\\\ \frac{d[C]}{dt} =0.9\frac{mol}{dm^3*s}\\\\\\ \frac{d[D]}{dt} =\frac{2}{-3} \frac{d[B]}{dt}=\frac{2}{-3}*-2.7\frac{mol}{dm^3*s} \\\\ \frac{d[D]}{dt} =1.8\frac{mol}{dm^3*s}[/tex]

Whereas the rates of consumption are negative and the rates of formation positive.

Best regards!

hope this will help you mate! :)

shadow

Answer:

C₄H₆O₃ + H₂O ⇒ 2C₂H₄O₂

Explanation:

Given:

C₄H₆O₃ + H₂O ⇒ C₂H₄O₂

Find:

Balance equation

Computation:

C₄H₆O₃ + H₂O ⇒ C₂H₄O₂

C₄H₆O₃ + H₂O ⇒ 2C₂H₄O₂

Number of atom should be equal in LHS and RHS.

So,

We have to multiply 2 in RHS to get equal number of atom.

Answer:

gravity exerts on it's mass brother

Answer:

For generating electricity, there must be a power source present and a total circuit. In order to generate electricity from citcrus fruit circuit, it is necessary to have materials like wire, a small bulb, lemon fruit, and two metal elements. The power source in this circuit is the lemon fruit. The metals that are taken here in this experiment are zinc and copper.

Explanation:

Answer: making sure the liquid inside the thermometer is at eye level when taking the temperature.

making sure the bulb of the thermometer does not touch the bottom of the beaker or the ice when taking the temperature

Explanation:

A thermometer is a device that's used to know the temperature of individuals. In order to ensure the accuracy of the readings in a thermometer, it is vital to making sure the liquid inside the thermometer is at eye level when taking the temperature.

Also, one should make sure the bulb of the thermometer does not touch the bottom of the beaker or the ice when taking the temperature.

The thermometer should not be used as a stirring rod. One should not shake it or play with it. Also, using either Celcius or Fahrenheit doesn't really matter.

Not every practice is accepted in temperature reading and these are the practices that apply are

Generally, The essential precautions that need to be taken when the using of thermometer encompass the following:

Do not use a thermometer as a stirring instrument,

Make positive that the liquid interior the thermometer is at eye stage when taking the temperature reading

Do not swing or shake down a thermometer reading

Therefore,the ways that apply are

Making sure the liquid inside the thermometer is at eye level when taking the temperature.

Making sure the bulb of the thermometer does not touch the bottom of the beaker or the ice when taking the temperature.

Option C and D are correct

For more information on  Temperature

https://brainly.com/question/13439286

Answer:

light, temperature, water, humidity, and nutrition.

Answer: It reacts vigorously with oxygen to form water.

Explanation:

Physical property is defined as the property of a substance which becomes evident during physical changes.

Example:  At room temperature its density is less than any other gas.

Liquefied [tex]H_2[/tex] boils at a very low temperature  [tex](-253^oC)[/tex]

Chemical property is defined as the property of a substance which becomes evident during chemical changes. These properties are defined when  a substance undergo changes in bonding of atoms.

Example:  It reacts vigorously with oxygen to form water.

Answer:

Following are the solution to the given points:

Explanation:

Oxalic acid volume [tex]= 25.00 \ mL = 0.0250 \ litres[/tex]  

KMnO4 volume [tex]= 12.70 \ ml = 0.0127 \ litres[/tex]

KMnO4 molarity [tex]= 0.0206\ M = 0.0206 \ \frac{mol}{l}[/tex]

In point a:

Its pink presence after full intake of oxalic acid with attachment to KMnO4 is suggested by the end-point of the process due to the small abundance of KMnO4, As just a self predictor, KMnO4 is used.

In point b:

[tex]H_2C_2O_4[/tex] molecules mole ratio to [tex]MnO_4^-[/tex] ions:  

The equilibrium for both the oxalic acid and KMnO4 reaction is suggested:

[tex]6H+ (aq) + 2MnO_4- (aq) + 5H_2C_2O_4 (aq) \rightarrow 10CO_2 (g) + 8H_2O (l) + 2Mn_2+ (aq)[/tex]

The reaction of 5 mol of oxalic acid is 2 mol [tex]MnO_4^-[/tex] ions  

[tex]H_2C_2O_4[/tex]: molecules mole proportion to [tex]MnO_4^-[/tex] ions:  

[tex]5 H_2C_2O_4[/tex]: : [tex]2MnO_4^-[/tex]

In point c:

The Moles of [tex]MnO_4^-[/tex] ions reacted with the [tex]H_2C_2O_4[/tex]:

The molar mass of the solution is the number of solute moles in each volume of water  

[tex]Molarity =\frac{moles}{Volume}\\\\Moles \ of\ KMnO_4 = Molarity \times volume[/tex]

Moles with ions reacted to mol with both the amount of : supplied.

In point d:

[tex]H_2C_2O_4[/tex] moles in the sample present:  

[tex]H_2C_2O_4[/tex] moles = moles [tex]MnO_4^-[/tex] ions [tex]\times[/tex] mole ratio  

[tex]H_2C_2O_4[/tex] moles in the sample = [tex]2.6162 \times 10^{-4}\ mol \times (\frac{5}{2})[/tex]

[tex]H_2C_2O_4[/tex] molecules = [tex]6,5405\times 10^{-4}[/tex] mol are present in the sample

In point e:

Oxalic acid molarity = [tex]\frac{mole}{volume}[/tex]  

                                  [tex]=\frac{ 6.54 \times 10^{-4} mol}{0.025\ L} \\\\ = 0.0260 \ M[/tex]

In point f:

Oxalic acid level by mass in the solution:  

Oxalic acid mass calculation:  

Oxalic acid molar weight = 90.0349 [tex]\frac{g}{mol}[/tex].  

Oxalic acid mass per liter = oxalic acid moles per liter [tex]\times[/tex] molar mass

                                          [tex]= 0.0260 \frac{mol}{L} \times 90.0349 \frac{g}{mol}\\\\= 2.3409 \frac{g}{L}\\\\ = 2.3409 \frac{g}{1000 \ mL}\\\\= 0.2409 \frac{g}{100 \ mL}[/tex]

When Oxalic acid solution density[tex]= 1.00 \ \frac{g}{mL}[/tex]  

Mass oxalic acid percentage = [tex]0.2409 \%[/tex]

Oxalic acid mass proportion [tex]= 0.24\% \ \frac{W}{v} \ \ Mass[/tex]

Answer:

8.77 grams

Explanation:

Answer: Its A or D                                                        

wish i had an actual answer sorry..  

The correct answer should be A

Answer:

Immunity

Explanation:

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if [tex]\rm O_2[/tex] is removed from the mixture. The reason is that collisions between [tex]\rm SO_2[/tex] molecules and [tex]\rm O_2\![/tex] molecules would become less frequent.

The reaction would not be at equilibrium for a while after [tex]\rm O_2[/tex] was taken out of the mixture.

Explanation:

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of [tex]\rm SO_3\, (g)[/tex] to the system, the backward reaction would have broken down the same amount of [tex]\rm SO_3\, (g)\![/tex]. So is the case for [tex]\rm SO_2\, (g)[/tex] and [tex]\rm O_2\, (g)[/tex].

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that [tex]\rm SO_2\, (g)[/tex] and [tex]\rm O_2\, (g)[/tex] molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield [tex]\rm SO_3\, (g)[/tex], only a fraction of the reactions will be fruitful.

Assume that [tex]\rm O_2\, (g)[/tex] molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer [tex]\!\rm O_2\, (g)[/tex] molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between [tex]\rm SO_2\, (g)[/tex] and [tex]\rm O_2\, (g)\![/tex] molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

The backward reaction rate is likely going to stay the same right after [tex]\rm O_2\, (g)[/tex] was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more [tex]\rm SO_3\, (g)[/tex] gets converted to [tex]\rm SO_2\, (g)[/tex] and [tex]\rm O_2\, (g)[/tex], the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

16.2 g of Water are formed

Answer:

Explanation:

Strong electrolytes completely ionize when dissolved, and no neutral molecules are formed in solution. Since NaCl is an ionic solid (s), which consists of cations Na+ and anions Cl−, no molecules of NaCl are present in NaCl solid or NaCl solution. The ionization is said to be complete.

Answer:

K3PO4(s) ----- 3K+(aq) + PO43–(aq)

Explanation:

edge 2021

Answer:

For neutral atoms, the number of valence electrons is equal to the atom's main group number. The main group number for an element can be found from its column on the periodic table. For example, carbon is in group 4 and has 4 valence electrons. Oxygen is in group 6 and has 6 valence electrons.

Explanation:

hope this help

Answer:

Water is water.

Explanation:

Water is a colorless and odorless substance found all over Earth. Water is made up of billions of molecules. ... Water on our planet flows as liquid in rivers, streams, and oceans; is solid as ice at the North and South Poles; and is gas (vapor) in the atmosphere. Water is also underground and inside plants and animals

Answer:

Settling and decanting is a method to reduce turbidity by letting the water sit for 2-24 hours so that the particulates settle to the bottom of the container. The clear water is then decanted off the top into a second container.

Explanation:

Settling and decanting is a method to reduce turbidity by letting the water sit for 2-24 hours so that the particulates settle to the bottom of the container. The clear water is then decanted off the top into a second container.

Answer:

because it starts to fall apart and it eventually just disappears like for example putting a bleach tablet it water it will take day to desolve but it desolves any way though.

Explanation:

Options A and B have wrong reasons because to feed on producers means that it is a primary consumer.

Option C is wrong because organism X only feeds on cornstalks.

Organism Y does feed on the grasshopper, which feeds on the cornstalks. It is a secondary consumer.

Hence Option D is correct.

Answer:

The answer is "[tex]2.73 \times 10^3 \ K[/tex] ".

Explanation:

Please find the complete question in the attachment.

The Formula for Ideal gas:

[tex]\to PV = nRT \\\\[/tex]

          [tex]= ( \frac{m}{M})RT[/tex]

[tex]\to Density\ \rho = \frac{m}{V} = \frac{PM}{RT}[/tex]

[tex]\to P= pressure\\\\\to V = volume\\\\ \to n = moles\ of \ gas \\\\\to R = molar \ gas \ constant\\\\ \to T = temperature\\\\ \to m = mass \\\\ \to M = molar \ mass[/tex]

[tex]\to P(Ar) = P(He) = 1.00 atm\\\\\to T(Ar) = ?\\\\ T(He) = 273.2 \ K\\\\\to M(Ar) = 39.948 \ \frac{g}{mol}\\\\ \to M(He) = 4.0026 \ \frac{g}{mol}\\\\\to \rho(Ar) = \rho(He)\\\\\bold{Formula: } \\\\ \to \frac{P(Ar)M(Ar)}{RT(Ar)} = \frac{P(He)M(He)}{RT(He)}\\\\\to \frac{1.00 \times 39.948}{(0.08206 \times T(Ar))} = \frac{1.00 \times 4.0026}{(0.08206 \times 273.2)}\\\\ \to T(Ar) = 2.73 \times 10^3 \ K[/tex]

Answer:

1. 1.21mol

2.8.5kg

Explanation:

the number of moles and grams of gas in the sample

Answer:

because he would've not known properly about it!

Answer:

3. 2.00 L of air at 100°C has twice the pressure of the same 2.00 L of air at  50°C

Explanation:

1)  doubling the pressure of a gas doubles its volume

This is incorrect because by increasing the pressure volume is going to be decreased.

2)  2.00 L of air at 227°C has twice the pressure of the same 2.00 L of air at  23°C

This is also incorrect because 2 L air at 227 °C will have the pressure more than twice time as compared to the 2L air at 23°C.

3. 2.00 L of air at 100°C has twice the pressure of the same 2.00 L of air at  50°C

This is correct;

According to Gay-Lussac Law,

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

Mathematical relationship:

P₁/T₁ = P₂/T₂

thus, by increasing the temperature pressure is also goes to increase in the same ratio.

4. halving the pressure of a gas halves its volume

This is incorrect because halving the pressure of gas increasing its volume.

Answer:

Liquid helium

Explanation:

Liquid helium is known to be the coldest element in the world.

It has a ridiculous boiling point of -452 degrees

It is considered the only element on earth that does not freeze to become solid. normally, it is usually in a form of cryogenic fluid or superfluid.

Hence, the coldest element is known as LIQUID HELIUM