If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)

Complete Question

A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 298 K. The final volume of the air in the tire is Vf= 1.50 L and the final pressure is Pf = 5.00 bar. Calculate the final temperature of the air in the tire.

Assume that Cv,m = 5R/2

Answer:

The value is [tex]T_f = 471.978 \ K[/tex]

Explanation:

From the question we are told that

The initial pressure is [tex]P_i = 1.00 \ bar[/tex]

The initial temperature is [tex]T_i = 298 K[/tex]

The final volume is [tex]V_f = 1.50 \ L[/tex]

The final pressure is [tex]P_f = 5.0 \ bar[/tex]

Generally the equation for reversible adiabatic compression is

[tex]P_i V_i^{\gamma} = P_f V_f^{\gamma} [/tex]

Generally from the ideal gas equation we have that

[tex]PV = n RT[/tex]

=> [tex] V = \frac{n RT}{P}[/tex]

So

[tex]P_i *[ \frac{n RT_i}{P_i}]^{\gamma} = P_f [ \frac{n RT_f}{P_f}]^{\gamma}[/tex]

=> [tex]P_i ^{1 - \gamma} * T_i ^{\gamma} = P_f ^{1-\gamma} * T_f^{\gamma}[/tex]

=> [tex]T_f ^{\gamma} = \frac{ P^{1 - \gamma * T_i^{\gamma}}}{ P_f ^{1 - \gamma}}[/tex]

=> [tex]T_f =[ \frac{ P_i ^{1 - \gamma} * T_i^{\gamma}}{P_f^{1 - \gamma}}]^{\frac{1}{\gamma} }[/tex]

=> [tex]T_f = T_i * [\frac{P_i}{P_f} ]^{\frac{1- \gamma}{\gamma}[/tex]

Here [tex]\gamma[/tex] is a constant mathematically represented as

[tex]\gamma = \frac{C_{P,m}}{C_{V, m}}[/tex]

Here

[tex]C_{P,m}} [/tex] is the molar heat capacity at constant pressure

and [tex]C_{V,m}} [/tex] is the molar heat capacity at constant volume given as

[tex]C_{V,m}} = \frac{5R}{2} [/tex]

Generally [tex]C_{P,m}} [/tex] for an ideal gas is mathematically represented as

[tex]C_{P,m}} = R +C_{V,m}} [/tex]

So

[tex]\gamma = \frac{\frac{5R}{2} + R}{ \frac{5R}{2} }[/tex]

=>   [tex]\gamma  =  \frac{7}{5} [/tex]

So

=> [tex]T_f  =  T_i *  [\frac{P_i}{P_f} ]^{\frac{1- \frac{7}{5}}{ \frac{7}{5}}[/tex]

=> [tex]T_f  = 298  *  [\frac{1}{5} ]^{\frac{1- \frac{7}{5}}{ \frac{7}{5}}[/tex]

=> [tex]T_f  = 471.978 \  K[/tex]

The Final temperature of the air in the tire = 471.978k

Given data :

Initial pressure ( [tex]P_{i}[/tex] ) = 1.00 bar

Initial temperature ( [tex]T_{i}[/tex] ) = 298 K

Final volume of air ( [tex]V_{f}[/tex] ) = 1.50 L

Final pressure ( [tex]P_{f}[/tex] ) = 5.00 bar

Assume Cvm = 5r / 2  ( missing data )

Determine the final temperature of the air in the tire

Applying the equation for a reversible Adiabatic compression and general ideal gas equation

[tex]Pi * [ \frac{nRTi}{Pi} ] \alpha = Pf * [ \frac{nRTf}{Pf}] \alpha[/tex]  ---- ( 1  )

Resolving equation ( 1 )

[tex]final temperature = Ti * [ \frac{Pi}{Pf}]^{\frac{1-\alpha }{\alpha } }[/tex]   ------- ( 2 )

where [tex]\alpha[/tex] ( mathematical constant ) = ( Cpm / Cvm ) --- ( 3 )

Cpm = molar heat capacity at constant pressure = R +  Cvm

Cvm = molar heat capacity at constant volume = 5r / 2

∴ Equation ( 3 ) becomes

    [tex]\alpha[/tex]  = [tex]\frac{7}{5}[/tex]

Final step : Determine the final temperature of the air in the tire

Input values back into equation ( 2 )

[tex]final temperature = 298 + [\frac{1}{5}] ^{\frac{1-\frac{7}{5} }{\frac{7}{5} } }[/tex]

     = 471.978 K

Hence we can conclude that the final temperature of the air in the tire is 471.978 K

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Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Given data

Average distance between nitrogen and oxygen atoms: 115 pm

Step 2: Convert the distance to meters (SI base unit)

We will use the conversion factor 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Convert the distance to millimeters

We will use the conversion factor 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

Answer:

a

Explanation:

Density is a physical property.

Answer:

You should rinse your head thoroughly ASAP, don't try to blow on the chemical or anything. Its going burn, you have to let your head cool off.

Explanation:

why rinse off your hand? you're getting the fluid away fast, so I can't infect your skin that much.

Answer:

Tvfbggbtb

Explanation:

Answer:

5.physical change

6.chemical change

7.physical change

8.conservation of mass

9.thermal energy

10.physical change

I honeslty dont know if this is right

explanation:

Answer:

123.0*12.35/(0.05*6.049)

first do in small bracket 123.0*12.35/0.30245

and then divide 123.0*40.83319557

and multiply 5022.483055

Given :

A 75.0 mL sample of water is heated to its boiling point.

To Find :

The heat (in kJ ) is required to vaporize it .

Solution :

Heat of vaporization of water , [tex]\Delta H_{vap}=40.65\ kJ/mol[/tex].

Molar mass of [tex]H_2O[/tex] , [tex]M=18\ g/mol[/tex] .

Density of water , [tex]\rho=1\ g/ml[/tex] .

Mass of 75 ml of water sample , [tex]m=75\ g[/tex] .

Moles of 75 g water = [tex]\dfrac{75}{18}=4.17\ mol[/tex] .

Now , heat required to vaporize 4.17 mol of water :

[tex]H=4.17\times 40.65\ kJ\\\\H=169.51\ kJ[/tex]

Therefore , heat required to boil 75 ml sample of water is 169.51 kJ.

Hence , this is the required solution .  

Answer:

An acid with Kg = 1.3 times 10-10.

Explanation:

The formula for pH of a buffer solution is as follows

pH = pKa + log [ A⁻ ] / [ HA ]

HA is acid concentration and A⁻ is concentration of conjugate base .

Generally the the second term is very small so neglecting this figure

pH = pKa

9.8 = pKa

If Ka = 10⁻¹⁰ x 1.3

pKa = - log ( 10⁻¹⁰ x 1.3 )

= 10 - log 1.3

= 9.8 approx .

Hence option F is correct .

Answer:

you use your sink and spray the little hose thingy and boom. or you can make it rain with mooney

Explanation:

no u can not make it rain in the house but if u want in the jar, here are the steps

Question: How can you make it rain indoors? 1) First, fill a glass jar with very hot tap water. 2) Then, flip the lid over and fill it with ice. ... In about 15 minutes you will see "rain" fall inside the jar .

hope this helped you

Answer:

20 mL

2 cL

Explanation:

We can use the base 10 decimal system to get equivalent measurements:

1 L = 1000 mL

1 L = 100 cL

Answer:

lithium phosphate

Explanation:

The name of the given ionic compound is lithium phosphate. The correct option is D.

Ionic compounds are neutral compounds composed of positively charged cations as well as negatively charged anions.

For binary ionic compounds, the identity of the cation is written first, accompanied by the title of the anion.

Ionic compounds are made up of ions that are held together through the force of attraction between oppositely charged ions.

One of the most well-known ionic compounds is common salt. Molecular compounds are made up of discrete molecules that are held together by sharing electrons, a process known as covalent bonding.

Lithium phosphate (Li3PO4) is an ionic compound that is primarily used as a conductive thin-film electrolyte in the fabrication of high performance energy storage devices.

Thus, the correct option is D.

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Answer:

Evaluate the result (Apex)

Explanation:

Answer:

If 33.6 g of NO and 26.9 g of O₂ react together, 51.52 grams of NO₂ can be formed.

Explanation:

You know: 2 NO (g) + O₂ (g) → 2 NO₂ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and product participate in the reaction:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the following amounts of reactant and product participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

You can apply the following rule of three to determine the limiting reagent: if by stoichiometry 60 g of NO react with 32 g of O₂, 33.6 g of NO with how much mass of O₂ will it react?

[tex]mass of O_{2}=\frac{33.6 grams of NO*32 grams of O_{2} }{60 grams of NO}[/tex]

mass of O₂= 17.92 grams

But 17.92 grams of O₂ are not available, there are 26.9 grams available. Since it has more mass than it needs to react with 33.6 grams of NO, NO will be the limiting reagent.

Then you apply the following rule of three: if by reaction stoichiometry 60 grams of NO form 92 grams of NO₂, 33.6 grams of NO how much mass of NO₂ does it form?

[tex]mass of NO_{2}=\frac{33.6 grams of NO*92 grams of NO_{2} }{60 grams of NO}[/tex]

mass of NO₂= 51.52 grams

If 33.6 g of NO and 26.9 g of O₂ react together, 51.52 grams of NO₂ can be formed.

Answer:friction

Explanation:

Answer:

Force, [tex]F=7.3\times 10^{36}\ N[/tex]

Explanation:

We need to find the force of the Earth on the moon be when the moon is 2.0D from the Earth.

The force that act between Earth and the Moon is gravitational force. It is given by :

[tex]F=\dfrac{Gm_1m_2}{r^2}[/tex]

We have,

[tex]m_1=\text{mass of Earth}=5.972 \times 10^{24}\ kg[/tex]

[tex]m_2=\text{mass of Moon}=7.34\times 10^{22}\ kg[/tex]

So,

[tex]F=\dfrac{6.67\times 10^{-11}\times 5.972 \times 10^{24}\times 7.34\times 10^{22}}{(2)^2}\\\\F=7.3\times 10^{36}\ N[/tex]

So, the force of the Earth on the Moon is [tex]7.3\times 10^{36}\ N[/tex].

Answer:

1050 mL

Explanation:

Step 1: Given data

Dosage of the liquid medication: 2.5 tablespoons (tbsp)

Times that the patient takes the medication: 4 times a day for a week

Step 2: Calculate the number of times the patient takes the medication

We will use the relationship 1 week = 7 days.

[tex]1 week \times \frac{7days}{1week} \times \frac{4dose}{1day} = 28 dose[/tex]

Step 3: Calculate the number of tablespoons

We will use the relationship 2.5 tbsp/1 dose.

[tex]28dose \times \frac{2.5tbsp}{1dose} = 70tbsp[/tex]

Step 3. Convert 70 tbsp to mL

We will use the relationship 1 tbsp = 15 mL.

[tex]70tbsp \times \frac{15mL}{1tbsp} = 1050 mL[/tex]

Answer:

[tex]Ka=5x10^{-4}[/tex]

[tex]pH=1.92[/tex]

Explanation:

Hello,

In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:

[tex]\% ionization=\frac{[H^+]}{[HA]}*100\%[/tex]

[tex][H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M[/tex]

Therefore the Ka is computed by using the equilibrium expression:

[tex]Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}[/tex]

And the pH:

[tex]pH=-log([H^+])=-log(0.012)\\\\pH=1.92[/tex]

Regards.

Answer: D

Explanation:

i just took the ck-12 quiz and that was the right answer

Answer:

increases reaction rates

Explanation:

Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

Answer:

1. more slowly than

2. larger than

3. weaker

Explanation:

Acetone molecules are bonded by very weak intermolecular forces when compared to that of the hydrogen bond between water molecules. This makes it very easy for acetone molecules to vaporize easily into its gaseous state, much more faster than water molecules (since the acetone molecules need a lesser amount of energy to break these bonds). Also, the boiling point of liquid acetone is much lower than that of water, meaning that it has a higher vapor pressure than that of water.

Answer:

A. True

Explanation:

This process usually cannot be reversed and will lead to the protein being permanently damaged.

Answer:

[tex]2C_{8}H_{18 (l)} + 25O_{2 (g)}[/tex] → [tex]16CO_{2(g)} +18H_{2}O_{(l)}[/tex]

Explanation:

Answer:

there are 4.0205 Litres of plasma in 8.5 pints

Explanation:

Recall that:

1 quart qt = 2 pints pt  and 1 quart = 946 mL

∴ two pints will contain 946 mL

and one pint will be = 946 mL/2

=  473 mL

Similarly, 8.5 pints will contain 473 mL × 8.5 = 4020.5  mL

We know that :

1000 mL = 1 Litres

Hence 4020.5 mL = (4020.5 /1000)L

= 4.0205 Litres

Therefore, there are 4.0205 Litres of plasma in 8.5 pints

There are 4.0205 Litres of plasma in 8.5 pints

1 quart qt = 2 pints pt  

and

1 quart = 946 mL

So, two pints will contain 946 mL

and one pint will be = [tex]946 mL\div 2[/tex]=  473 mL

Now

8.5 pints will contain

[tex]= 473 mL \times 8.5[/tex]

= 4020.5  mL

We know that :

1000 mL = 1 Litres

Hence 4020.5 mL = (4020.5 /1000)L

= 4.0205 Litres

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Answer:

CH3OH > CH3Cl > CH4

Explanation:

The boiling point of a compound depends on the nature and magnitude of intermolecular forces acting between its molecules.

Methanol contains the -OH group, hence hydrogen bonds are dominant intermolecular forces in the compound. Presence of hydrogen bonds increases the boiling point of a substance, hence methanol has the highest boiling point followed by chloromethane which has dipole-dipole interaction as dominant intermolecular forces and methane having only weak dispersion forces as dominant intermolecular forces.

Answer:

5.265L/body ; 68.9×10³ kg / beat

Explanation:

Given the following :

Resting pulse rate = 81 beats / minute

Amount of blood pumped per beat = 65mL = (65 / 1000)L = 0.65L

Density of blood = 1060 kg/m^3

Circulation time = 1 minute for a person at rest

A.) How much blood is in the body?

Resting pulse rate × amount of blood pumped per beat * circulating time

(81 beat / min) × 0.65L/beat * 1minute / body = 5.265L / body

B.) b. On average, what mass of blood does the heart pump with each heart beat?

0.065L/beat * 1m³/1000L= 0.000065 m³ / beat

From density = mass / volume ;

Average mass = density * volume

= 1060kg/m³ * 0.000065 m³ / beat

= 0.0689kg / beat

= 68.9 × 10³ kg / beat

Answer:

Following are the explanation of the Rube Goldberg device:

Explanation:

According to the Rube Goldberg devices, which conform with "the energy conservation law," choose a chain of events to carry out such a basic task differently, for this unit, a range of instant theatrical resources are converted into possible energy. It is also responsive to an energy conservation law.