During the winter months, many locations experience snow and ice storms. It is a common practice to treat roadways and sidewalks with salt, such as NaCl . If a 11.3 kg bag of NaCl is used to treat the sidewalk, how many moles of NaCl does this bag contain

Answers

Answer 1

Answer:The moles of NaCL in the 11.3kg bag  is   193.36moles

Explanation:

Given that  a bag of NaCl = 11.3kg

1kg = 1000g

therefore 11.3 kg = 11,300g

Remember that

No of moles = mass of subatance/ molar mass of substance

The molar mass of NaCl = Na + Cl= 22.989769  + 35.453 =58.442769≈ 58.44g/mol

No of moles = mass of subatance/ molar mass of substance

                    = 11300g/ 58.44g/mol  =  193.36 moles

The moles of NaCL in the 11.3kg bag  is   193.36 moles.


Related Questions

Convert 150 K to degrees C.

Answers

Answer:

K = 150, C = - 123.15°

Explanation:

Kelvin = Celcius + 273.15 / 0 Kelvin = - 273.14 C

_____________________________________

Thus,

150 K = Celcius + 273.15,

150 - 273.15 = C,

C = -123.15 degrees

Solution, C = - 123.15°

Answer:

C=-123.15

Explanation:

This is easy

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g) 2.10 g were obtained. What is the percent yield of the reaction?

Answers

Answer:

[tex]Y=58.15\%[/tex]

Explanation:

Hello,

For the given chemical reaction:

[tex]Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)[/tex]

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

[tex]n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3[/tex]

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

[tex]n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3[/tex]

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

[tex]m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3[/tex]

Finally, we compute the percent yield with the obtained 2.10 g:

[tex]Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%[/tex]

Best regards.

A sample of chlorine gas starting at 686 mm Hg is placed under a pressure of 991 mm Hg and reduced to a volume of 507.6 mL. What was the initial volume of the chlorine gas container if the process was performed at constant temperature

Answers

Answer:

The initial volume of the chlorine gas [tex]V1=733.28mL[/tex]

Explanation:

Given:

P1= 686mmHg

P2= 991mmHg

V2= 5076mL

V1=?

According to Boyle's law which states that at a constant temperature, the pressure on a gas increases as it's volume decreases.

It can be expressed as : P1V1 = P2V2

Where P1 is the initial pressure

P2= final pressure

V1= initial volume

V2 = final volume

[tex]V1= (P2V2)/P1[/tex]

V1= (991mmHg*507.6mL)/686mmHg

V1=503031.6/686

[tex]V1=733.28mL[/tex]

Therefore, The initial volume of the chlorine gas [tex]V1=733.28mL[/tex]

A solid is dissolved in a liquid, and over time a solid forms again. How can
you confirm the type of change that took place?
A. Testing the new solid to show that its properties are the same as
the starting solid would confirm that a physical change took
place.
B. The solid dissolving in a liquid is confirmation that a chemical
change took place.
C. The solid forming from the liquid is confirmation that a physical
change took place.
D. Showing that the total mass of the solid and liquid changed would
confirm that a chemical change took place.

Answers

I think B is write but even I’m not sure

When an automobile engine starts, the metal parts immediately begin to absorb heat released during the combustion of gasoline. How much heat will be absorbed by a 165 kg iron engine block as the temperature rises from 15.7°C to 95.7°C? (The specific heat of iron is 0.489 J/g·°C.)

Answers

Answer:

H = 4,034,250 J

Explanation:

Mass, m = 165kg = 165,000g (Converting to grams)

Initial temperature = 15.7°C

Final temperature = 95.7°C

Temperature change, ΔT = 95.7 - 15.7 = 50°C

Specific heat capacity, c = 0.489 J/g·°C

Heat = ?

All the parameters are related with the equation below;

H = m * c * ΔT

H = 165000 * 0.489  * 50

H = 4,034,250 J

Because cylinder with a volume of 3.00dm^3 contains 8.00 moles of oxygen gas at a temperature of 50.0 K. What is the pressure inside the cylinder? ( R=8.31)

Answers

Answer:

Option A. 1110 KPa.

Explanation:

The following data were obtained from the question:

Volume (V) = 3 dm³

Number of mole (n) = 8 moles

Temperature (T) = 50K

Gas constant (R) = 8.31 KPa.dm³/Kmol

Pressure (P) =..?

Pressure inside the cylinder can be obtained by using the ideal gas equation as follow:

PV =nRT

P x 3 = 8 x 8.31 x 50

Divide both side by 3

P= (8 x 8.31 x 50) /3

P = 1108 ≈ 1110 KPa

Therefore, the pressure inside the cylinder is 1110 KPa

Nitrogen monoxide is produced by combustion in an automobile engine. For the following reaction, 4.36 grams of nitrogen monoxide are mixed with excess oxygen gas . The reaction yields 5.46 grams of nitrogen dioxide . nitrogen monoxide ( g ) oxygen ( g ) nitrogen dioxide ( g ) What is the theoretical yield of nitrogen dioxide

Answers

Answer:

[tex]6.68~g~NO_2[/tex]

Explanation:

We have to start with the combustion reaction:

[tex]NO~+~O_2~->~NO_2[/tex]

Then we can balance the reaction:

[tex]2NO~+~O_2~->~2NO_2[/tex]

If we want to find the theoretical yield, we have to calculate the amount of [tex]NO_2[/tex]. To do this, we have to first convert the 4.36 g of [tex]NO[/tex] to moles [tex]NO[/tex] (using the molar mass 30 g/mol), then we have to convert from moles of [tex]NO[/tex] to moles of [tex]NO_2[/tex] (using the molar ratio) finally, we have to convert from moles of [tex]NO_2[/tex] to grams of [tex]NO_2[/tex] (using the molas mass 46 g/mol), so:

[tex]4.36~g~NO\frac{1~mol~NO}{4.36~g~NO}\frac{2~mol~NO_2}{2~mol~NO}\frac{46~g~NO_2}{1~mol~NO_2}=6.68~g~NO_2[/tex]

I hope it helps!

Perform the conversions between energy units.
861 kJ=
J
3495 kcal=
kJ
7.84×106

Answers

Answer:

861 kJ = 861000 J,

3495 kcal = 14623.08 kJ

Explanation:

As I mentioned before, the last bit " 7.84 × 106 " just threw me off track, so I am simply going to assume that that does not appear in your question.

_______________________________________________________

Now we have 861 kilojoules, and have to convert it into joules for this first bit. Kilo being equal to 1000, to convert to joules you would have to multiply 861 by 1000, = 861000 Joules.

This second bit here asks us to convert 3495 kilocalorie to kilojoules. The difference between the two is that one is about 4.18 times greater than the other, so 3495 kilocalorie = 3495 * 4.18 = 14623.08 kilojoules.

Hope that helps!

What is the coefficient for oxygen in the balanced equation? C 5H 12 + ? O2 → ? CO2 + ? H2O. 2 4 5 6 8

Answers

Answer:

8

Explanation:

When you balance the entire equation, you should get:

C5H12 + 8O2 ---> 5CO2 + 6H2O

The number of atoms or molecules whose concentration determine the rate of a reaction is called

Answers

order of the reaction

The element silver has an atomic weight of 108 and consists of two stable isotopes silver-107 and silver-109. The isotope silver-107 has a mass of 107 amu and a percent natural abundance of 51.8 %. The isotope silver-109 has a percent natural abundance of 48.2 %. What is the mass of silver-109

Answers

Answer:

109

Explanation:

Let silver-107 be isotope A

Let silver-109 be isotope B

Let silver-107 abundance be A%

Let silver-109 abundance be B%

The following data were obtained from the question:

Atomic weight of silver = 108

Mass of isotope A (silver-107) = 107

Abundance of isotope A (silver-107) = 51.8%

Abundance of isotope B (silver-109) = 48.2%

Mass of isotope B (silver-109) =?

Now, we shall determine the mass silver-109 as follow:

Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

108 = [(107 x 51.8)/100] + [(Mass of B x 48.2)/100]

108 = 55.426 + (Mass of B x 0.482)

Collect like terms

Mass of B x 0.482 = 108 – 55.426

Mass of B x 0.482 = 52.574

Divide both side by 0.482

Mass of B = 52.574/0.482

Mass of B = 109

Therefore, the mass of silver-109 is 109.

1. Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estimating their corresponding hydronium ion concentrations ([H3O+] using the ion product constant of water (Kw).
Kw = [H3O+][OH−] = [1×10−7 M][1×10−7 M] = 1×10−14 M
Hydronium ion concentration [H3O+] Solution condition
Greater than 1×10−7 M Acidic
Equal to 1×10−7 M Neutral
Less than 1×10−7 M Basic
Drag the appropriate items to their respective bins.
1. [OH−] = 6×10−12 M
2. [OH−] = 9×10−9 M
3. [OH−] = 8×10−10 M
4. [OH−] = 7×10−13 M
5. [OH−] = 2×10−2 M
6. [OH−] = 9×10−4 M
7. [OH−] = 5×10−5 M
8. [OH−] = 1×10−7 M
A. Acidic
B. Neutral
C. Basic
2. A solution has [H3O+] = 5.2×10−5M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [OH−] of the solution.
3. A solution has [OH−] = 2.7×10−2M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [H3O+] of the solution.

Answers

Answer:

Question 1.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

[H₃O⁺] = 3.70 * 10⁻¹¹ M

Explanation:

The ion product constant of water  Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M² is a constant which gives the product of the concentrations of hydronium and hydroxide ions of dissociated pure water. The concentrations of the two ions are both equal to 1 * 10⁻⁷ in pure water.

A solution that has [OH⁻] greater than 1 * 10⁻⁷ is basic while one with [OH⁻] less than 1 * 10⁻⁷ is acidic.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[OH⁻] = 1 * 10⁻¹⁴ M²/ [H₃O⁺]

[OH⁻] = 1 * 10⁻¹⁴ M²/5.2*10⁻⁵ M

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ [OH⁻]

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ 2.7 * 10⁻² M

[H₃O⁺] = 3.70 * 10⁻¹¹ M

What is an ion?
A. An atom that has lost or gained 1 or more electrons
O B. An atom that has lost or gained 1 or more neutrons
O C. An atom that has lost or gained 1 or more protons
D. An atom that differs in mass from another atom of the same
element

Answers

Answer:

An ion is an atom that has lost or gained one or more electrons.

Explanation:

Ions are positively or negatively charged atoms of elements. This is because they can give, take, or share electrons with other elements to encourage the formation of chemical bonds.

Protons are what decide the chemical identity of the element. So, for example, if an atom has 11 protons, we know that will be a Sodium (Na) atom. A loss or gain of protons completely changes the chemical identity of the element and it will then become another element.

Electrons are what give an atom a neutral electrical charge (if that atom has the number of protons and neutrons normally described for the element - otherwise, a discrepancy or gain in neutrons is referred to as an isotope and declares that ions have nothing to do with the mass of an element).

With this information, you can realize that neutrons and protons have nothing to do with ions and you can confirm that ions are atoms that have lost or gained one or more electrons.

The reason for the dramatic decline in the number of measles cases from the 1960s to 2010 in the United States was because the vaccine

Answers

Answer:

It was because the vaccine generated actively acquired immunity, that is, inoculation of a portion of the measles virus so that the body forms the antibodies for a second contact and thus can destroy it without triggering the pathology.

Explanation:

Vaccines are methods of active acquired immunity since the antibody is not passively inoculated, it is manufactured by the body with a physiological process once part of the virus is inoculated.

The measles virus most of all affected the lives of infants or newborn children with severe rashes and high fevers that led to death.


An atorn has 9 electrons and 9 protons at the start. If it loses 2 electrons, the net charge on the atom will be. If the atom instead
gains 4 electrons, the net charge will be

Answers

Answer:

If it loses 2 electrons, the net charge on the atom will be 2+

If the atom instead gains 4 electrons, the net charge will be 4-

Explanation:

It is based on adding and subtracting charges. Protons are +1 and electrons are -1

If the atom has 9 protons and 9 electrons, the net charge is +9 + (-9) = 0. The +9 is the 9 protons and the -9 is from the 9 electrons.

If two electrons are taken away, there would be 9-2 or 7 electrons with 9 protons. The net charge would then be +9 + (-7) = +2. +9 comes from the 9 electrons and -7 is from the 7 electrons.

So, if two electrons are taken away, the net charge is +2.

Similarly, if the atom gains 4 electrons, there will be 9+4 or 12 electrons and 9 protons. The net charge would then be +9 + (-12) = -4. +9 comes from the 9 protons and -12 comes from the 12 electrons.

So, if 4 electrons are added, the net charge is -4.

Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2? 7. How many grams of ClF3 form from 130.0 grams of Cl2 when F2 is in excess? 8. How many grams of F2 are needed to react with 3.50 grams of Cl2?

Answers

Answer:

5) 1.147 moles Cl2

6) 12.57 grams ClF3

7)  339.10 grams ClF3

8) 5.63 grams F2

Explanation:

Step 1: Data given

Number of moles F2 = 3.44 moles

Molar mass F2 = 38.00 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 3.44 moles F2 we'll need 3.44/3 = 1.147 moles Cl2

Step 1: Data given

Number of moles F2 = 0.204 moles

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles ClF3

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 0.204 moles F2 we'll have 2/3 * 0.204 = 0.136 moles

Step 4: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 0.136 moles * 92.448 g/mol

Mass ClF3 = 12.57 grams ClF3

Step 1: Data given

Mass of Cl2 = 130.0 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 130.0 grams / 70.9 g/mol

Moles Cl2 = 1.834 moles

Step 4: Calculate moles

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 1.834 moles Cl2 e'll have 2*1.834 = 3.668 moles ClF3

Step 5: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 3.668 moles * 92.448 g/mol

Mass ClF3 = 339.10 grams ClF3

Step 1: Data given

Mass of Cl2 = 3.50 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = Mass Cl2 / molar mass Cl2

Moles Cl2 = 3.50 grams / 70.9 g/mol

Moles Cl2 = 0.0494  moles

Step 4: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2

For 0.0494 moles we need 3*0.0494 = 0.1482 moles

Step 5: Calculate mass F2

Mass F2 = moles F2 * molar mass F2

Mass F2 = 0.1482 moles * 38.00 g/mol

Mass F2 = 5.63 grams F2

Q4. A 0.100 M solution of the weak acid HA was titrated with 0.100 M NaOH.The pH
measured when Vb = ½ Ve was 4.62. Using activity coefficients,calculate pKa.The size of the
A− anion is 450 pm.​

Answers

what grade are u in nnnn

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a balanced net ionic equation to show why the solubility of Ni(OH)2 (s) increases in the presence of cyanide ions and calculate the equilibrium constant for this reaction. For Ni(CN)42- , Kf = 1.0×1031 . Use the pull-down boxes to specify states such as (aq) or (s).

Answers

Answer: Equilibrium constant for this reaction is [tex]2.8 \times 10^{15}[/tex].

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

[tex]Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)[/tex]

We know that,

      K = [tex]K_{f} \times K_{sp}[/tex]

We are given that, [tex]K_{f} = 1.0 \times 10^{31}[/tex]

and,    [tex]K_{sp} = 2.8 \times 10^{-16}[/tex]

Hence, we will calculate the value of K as follows.

     K = [tex]K_{f} \times K_{sp}[/tex]

     K = [tex](1.0 \times 10^{31}) \times (2.8 \times 10^{-16})[/tex]

        = [tex]2.8 \times 10^{15}[/tex]

Thus, we can conclude that equilibrium constant for this reaction is [tex]2.8 \times 10^{15}[/tex].

In the presence of a strong base, the following reaction between (CH3)3CCl and OH- occurs: (CH3)3CCl + OH- → (CH3)3COH + Cl- Studies have suggested that the mechanism for the reaction takes place in 2 steps: Step 1) (CH3)3CCl → (CH3)3C+ + Cl- (slow) Step 2) (CH3)3C+ + OH- → (CH3)3COH (fast) What is the rate law expression for the overall reaction? Group of answer choices

Answers

Answer:

D. rate = k [(CH3)3CCl]

Explanation:

(CH3)3CCl + OH- → (CH3)3COH + Cl-

The mechanisms are;

Step 1)

(CH3)3CCl → (CH3)3C+ + Cl- (slow)

Step 2)

(CH3)3C+ + OH- → (CH3)3COH (fast)

In kinetics, the slowest step is the ratee determining step.

For a given reaction;

A → B + C, the rate law expression is given as;

rate = k [A]

In this problem, from step 1. The rate expression is;

rate = k [(CH3)3CCl]

Among three bases, X−, Y−, and Z−, the strongest one is Y−, and the weakest one is Z−. Rank their conjugate acids, HX, HY, and HZ, in order of decreasing strength. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them.

Answers

Answer: HZ > HX > HY in order of decreasing strengths.

Explanation: Generally, the rule is that the stronger the acid, the weaker its conjugate base and vice versa; same rule applies for bases and their conjugate acids.

So the weakest base Z- would have the strongest conjugate acid. Consequently, the strongest base Y- would have the weakest conjugate acid.

I hope this was MORE helpful as this is the correct answer.

The ranking of the conjugate acids in order of decreasing strength (i.e from strongest to weakest) is; HZ < HX < HY

First we must know that the stronger a base is, the weaker is it's conjugate acid and the weaker a base is, the stronger is it's conjugate acid.

Therefore, the order of decreasing strength of the conjugate acid is; HZ < HX < HY

Read more:

https://brainly.com/question/23917439

What force is needed to accelerate a truck with mass 2,000 kg at a rate of 5.0
m/s22
O A. 1,000 N
B. 20,000 N
C. 2,000 N
O D, 10,000 14

Answers

Answer:

D

Explanation:

[tex]F=ma \\\\F=2000\cdot 5=10,000N[/tex]

Hope this helps!

The lock-and-key model and the induced-fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Following are several statements concerning enzyme and substrate interaction. Indicate whether each statement is part of the lock-and-key model, the induced-fit model, or is common to both models.

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions

Answers

Answer:

The lock-and-key model:

c. Enzyme active site has a rigid structure complementary

The induced-fit model:

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.

Common to both The lock-and-key model and The induced-fit model:

b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.

d. Substrate binds to the enzyme through non-covalent interactions

Explanation:

Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.

The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.

The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.

an auto of an element has 17 protons in its nucleus.a) write the electronic configuration of the atom.b)to what period and group does the element belong​

Answers

Answer:

i hope it will help you

Explanation:

electronic configuration 1s²,2s,²2p^6,3s²3p^6,4s^1

as it has one electron in its valence shell so it is the member of group 1A(ALKALI METALS)  and the number of shells is 4 so it is in period 4

what type of matter is toluene

Answers

Answer:

is an organic chemical conpond

The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC2O4. The CaC2O4 weighed 40.3 g. What is the percent of calcium in the original sample

Answers

Answer:

128 gram of CaC2O4 contains 40 gram of Calcium

40.3 gram of CaC2O4 cotnains = 40*40.3/128 = 12.59 gram of Calcium

out of 15 gram 12.59 gram is Calcaim that means around 50% of orginal sample has Calcium

A certain lightbulb containing argon at 1.20 atm and 18°C is
heated to 85°C at constant volume. Calculate its final pressure
(in atm).

Answers

Answer:

certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P 1 T 1 P 2 T 2 ... Ideal Gas Equation 5.4 Charles' law: V T (at constant n and P ) ... Consider a case in which two gases, A and B , are in a container of volume V.

Explanation:

In which of the following reactions will Kc = Kp? a. 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g) b. SO3(g) + NO(g) ⇌ SO2(g) + NO2(g) c. 2 N2(g) + O2(g) ⇌ 2 N2O(g) d. 2 SO2(g) + O2(g) ⇌ 2 SO3(g)

Answers

Answer:

The correct option is b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)

Explanation:

The relation between Kc and Kp is given by the following equation:

[tex]Kp = Kc (RT)^{dn}[/tex]

where R is the gas constant (0,082 L.atm/K.mol), T is the temperature (in K) and dn is the change in moles.

The change in moles (dn) is calculated as:

dn = moles of products - moles reactants

If dn=0, RT= 1 ⇒ Kc=Kp

We calculate dn for each reaction from the estequiometrial coefficients of products and reactants as follows:

a) 4 NH₃(g) + 3 O₂(g) ⇌ 2 N₂(g) + 6 H₂O(g)

dn= (2+6) - (4+3) = 1 ⇒ Kc ≠ Kp

b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)

dn = (1+1) - (1+1)= 0 ⇒ Kc = Kp

c) 2 N₂(g) + O₂(g) ⇌ 2 N₂O(g)

dn= 2 - (2+1) = -1 ⇒ Kc ≠ Kp

d) 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)

dn = 2 - (2+1) = -1 ⇒ Kc ≠ Kp

The reaction in which Kc=Kp is b), because reactants and products have the same number of moles.

A friend asks you to help them decide which crackers are healthier. Comparing approximately equal serving sizes of 1 cracker (approx. 4.5 g Breton serving size vs. 4.7 g Triscuit), which would be a better choice with regards to calories, fat and sodium content?

a. Breton
b. Triscsuit

Answers

Answer:

The correct option is "Triscsuit"

Explanation:

In my opinion the correct option is tricsuit, because it has 0% saturated fat and TRANS fat, which is healthy fats since these fats are the worst for our body.

They also contain sodium but their levels are not high enough to trigger hypertension.

Substances like krypton, which is a gas at room temperature and pressure, can often be liquified or solidified only at very low temperatures. At a pressure of 1 atm, does not condense to a liquid until –153.2°C and does not freeze until –157.1°C. What are the equivalent absolute temperatures?

Answers

Answer:

The boiling and freezing temperatures of krypton at absolute scale are 119.95 K and 116.05 K, respectively.

Explanation:

The absolute temperature on SI units corresponds to Kelvin scale, whose conversion formula in terms of the Celsius scale is:

[tex]T_{K} = T_{C} + 273.15[/tex]

Where:

[tex]T_{K}[/tex] - Absolute temperature, measured in Kelvins.

[tex]T_{C}[/tex] - Relative temperature, measured in Celsius.

Finally, freezing and boiling temperatures are converted into absolute scale:

Boiling temperature

[tex]T_{K} = (-153.2 + 273.15)\,K[/tex]

[tex]T_{K} = 119.95\,K[/tex]

Freezing temperature

[tex]T_{K} = (-157.1 + 273.15)\,K[/tex]

[tex]T_{K} = 116.05\,K[/tex]

The boiling and freezing temperatures of krypton at absolute scale are 119.95 K and 116.05 K, respectively.

The partial Lewis structure that follows is for a hydrocarbon molecule. In the full Lewis structure, each carbon atom satisfies the octet rule, and there are no unshared electron pairs in the molecule. The carbon-carbon bonds are labeled 1, 2, and 3.
A) How many hydrogen atoms are in the molecule?
B) Rank the carbon-carbon bonds in order of increasing bond length.
C) Which carbon-carbon bond is the strongest one?

Answers

Answer:

A) How many hydrogen atoms are in the molecule?

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