A sample of nickel is heated to 99.8 degrees C and placed in a coffee cup calorimeter containing 150.0g water at 23.5 degrees C. After the metal cools, the final temperature of metal andwater mixture is 25.0 degrees C. If the specific heat capacity of nickel is 0.444J/(degreesCg),
what mass of nickel was originally heated?

Answer: a contest in which you are eliminated if you fail to spell a word correctly.

Answer:

1122.5 mL

Explanation:

In the first scenario, by Dalton's law, the total pressure is the sum of the partial pressures of the components. So, the partial pressure of the gas is:

P1 = Ptotal - Pwater = 742 - 36 = 706 torr

By the ideal gas law, the change in a state of a gas can be calculated by:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is the temperature in K, 1 is the state 1, and 2 the state 2.

P1 = 706 torr, V1 = 1350 mL, T1 = 32ºC + 273 = 305K

P2 = 760 torr, T2 = 0ºC + 273 = 273 K

706*1350/305 = 760*V2/273

760V2/273 = 3124.92

760V2 = 853102.62

V2 = 1122.5 mL

The density of sulfur hexafluoride gas at 15°C and 1 atm is approximately 6.52 g/L. This calculation is done using the ideal gas law and assuming sulfur hexafluoride behaves as an ideal gas under these conditions.

To calculate the density of sulfur hexafluoride gas at 15° C and 1 atm, we can use the ideal gas law. The ideal gas law states PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin. First, convert 15°C to Kelvin by adding 273.15, which gives 288.15 K. The molar mass of sulfur hexafluoride is around 146.06 g/mol. R is 0.0821 L·atm/(K·mol), and P is 1 atm. So, you can solve for density (d) which is mass/volume, or essentially (n*M)/V.

By rearranging the ideal gas law, V=nRT/P, and substituting into the density equation, we get d=P*M/(R*T) = (1 atm * 146.06g/mol) / (0.0821 L·atm/mol·K * 288.15 K) = approximately 6.52 g/L to three significant figures.

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The density of sulfur hexafluoride gas at 15°C and 1atm, assuming it behaves as an ideal gas, is approximately 6.52g/L.

The question asks for the density of sulfur hexafluoride gas at exactly 15°C and 1atm. We can calculate this by using the ideal gas law which states PV=nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin. We rearrange this to n/V=P/RT which gives us the molar density. From the molar density, we can easily determine the density by multiplying it by the molar mass. To convert 15°C to Kelvin, we add 273.15 to give us 288.15K. This gives us a molar density of 1atm/(0.0821 L.atm/K.mol * 288.15K) = 0.0446 mol/L. The molar mass of sulfur hexafluoride (SF6) is about 146g/mol. Multiply this by the molar volume gives us a density of 6.52g/L, to three significant figures.

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Answer:

Explanation:

It is possible to obtain ΔG of a reaction using:

ΔG = ΔG° + RT ln Q

Where:

ΔG° is standard change in Gibbs free energy (-32,8 kJ/mol)

R is gas constant (8,314472x10⁻³ kJ/molK)

T is temperature (298 K)

And Q is reaction quotient. For the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Q is: [NH₃]² /[H₂]³[N₂]

Replacing:

ΔG = -32,8kJ/mol + 8,314472x10⁻³ kJ/molK×298 K ln [0,6]² /[0,1]³[0,1]

ΔG = -12,5 kJ/mol

I hope it helps!

The change in Gibbs free energy (ΔG) for the reaction N2(g) + 3H2(g) --> 2NH3(g) at 298 K with partial pressures Pn2= 0.1 atm, Ph2=.1 atm and PNH3=.6 atm, is about -29.6 kJ/mol.

To calculate the Gibbs free energy change under nonstandard conditions, we should use the equation ΔG = ΔG° + RTlnQ, where ΔG° is the standard change in Gibbs free energy, R is the universal gas constant (in this case, 8.314 J/(mol·K)), T is the temperature in Kelvin, and Q is the reaction quotient.

Given that ΔG° = -32.8 kJ/mol, R = 8.314 J/(mol·K), and T = 298 K, we find the reaction quotient Q by substituting the partial pressures of the substances into the balanced chemical equation: Q = [NH3]² / ([N₂][H₂]³) = (0.6)² / ((0.1)(0.1)³) = 360.  

Then, you substitute these values into the ΔG equation, making sure to convert ΔG° and the result of RTlnQ to the same units (either J or kJ): ΔG = ΔG° + RTlnQ = -32.8 kJ/mol + 8.314 J/(mol·K) x 298 K x ln(360), which gives ΔG = about -29.6 kJ/mol. The negative value indicates that the reaction is spontaneous under these conditions.

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In the substitution reaction of P-fluoroanisole with sulfur trioxide and sulfuric acid, the major product is P-fluoroanisole-sulfonic acid. This is an example of an Electrophilic Aromatic Substitution reaction.

The question involves a reaction of P-fluoroanisole with sulfur trioxide and sulfuric acid. This is a substitution reaction that happens under the influence of a strong acid like sulfuric acid.

In this reaction, sulfur trioxide (SO3) reacts with P-fluoroanisole to replace a hydrogen atom with a sulfonic acid group (–SO3H), forming P-fluoroanisole-sulfonic acid as the major product of the reaction. This type of reaction is a part of a broader class of reactions known as Electrophilic Aromatic Substitution reactions.

However, it is important to note that drawing the resulting chemical structure requires knowledge of organic chemistry and its conventions. Given the complexity of this information, a written description may not fully capture the details, and it is recommended to refer to a textbook or online resource where visualizations of such substitutions are possible.

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The pH of the solution after the titration of 0.10M KOH solution with 0.10M HCl is 7. This is due to complete neutralization of the acid and base, leaving water which has a pH of 7 at 25 degrees Celsius.

The question involves a titration process of KOH solution with HCl solution. KOH and HCl are a strong base and a strong acid respectively, and they react in stoichiometrically equal amounts. Since the molarities and amount of KOH and HCl are the same, it means all the KOH and HCl react completely, leaving no excess.

The result of the reaction is water (H2O) and a salt (KCl), and since KCl does not hydrolyze, the only contribution to the pH of the solution will be from the water itself. The pH of pure water is 7 at 25 degrees Celsius because it is neutral (neither acidic nor basic at this temperature).

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Answer:

The halogens located in group 17

Explanation:

The change in energy of a neutral atom when an electron is added is known as electron affinity. Fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At) are known as halogens and conform the group 17 of the periodic table, the group that presents higher electron affinity. This characteristic happens because of their atomic structures, it is easier for them to gain electrons because they are closer to form octets.

Here I present you halogens' electron affinities values:

The use of negative signs is to indicate the RELEASE of energy.

I hope you find this information useful and interesting! Good luck!

Answer:

The sample of indium, has an average atomic mass of 114.818 amu

Explanation:

Step 1: Data given

Indium is made of 95.7 % 115In and 4.3% 113In

113In has an atomic mass of 112.904 amu

115In has an atomic mass of 114.904amu

Step 2: Calculate the average atomic mass

The average atomic mass = X

0.957* 114.904 + 0.043* 112.904 = X

109.963 + 4.855 = X

X = 114.818

The sample of indium, has an average atomic mass of 114.818 amu

Rounded to the nearest tenth, the average atomic mass of indium is approximately 114.9 amu. So, the correct options provided is:C

To calculate the average atomic mass of a sample of indium, you can use the following formula:

Average Atomic Mass = (% abundance of isotope 1 × atomic mass of isotope 1) + (% abundance of isotope 2 × atomic mass of isotope 2) + ...

In this case, you have two isotopes of indium, 115In and 113In, with the following information:

- % abundance of 115In = 95.7%

- % abundance of 113In = 4.3%

- Atomic mass of 115In = 114.904 amu

- Atomic mass of 113In = 112.904 amu

Plug these values into the formula:

Average Atomic Mass = (0.957 × 114.904 amu) + (0.043 × 112.904 amu)

Calculate each part:

Average Atomic Mass = (110.009648 amu) + (4.855672 amu)

Now, add these values together:

Average Atomic Mass = 114.86532 amu

Rounded to the nearest tenth, the average atomic mass of indium is approximately 114.9 amu. So, the correct options provided is:C

To know more about atomic mass :

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Answer:

3.41 x10⁶ torr

Explanation:

To solve this problem we need to remember the equivalency:

1 torr = 133.322 Pa

Then we can proceed to convert 4.55×10⁸ Pa into torr. To do that we just need to multiply that value by a fraction number, putting the unit that we want to convert from in the denominator, and the value we want to convert to in the numerator:

4.55x10⁸ Pa * [tex]\frac{1torr}{133.322Pa} =[/tex] 3.41 x10⁶ torr

Answer:

The reaction that will occur are:

[tex]Mn(s)+NiCl_{2}(aq)->MnCl_{2}(aq)+Ni(s)[/tex]

Explanation:

In the activity series, a higher metal can displace a lower metal from its salt.

Mn is placed in a higher position than Ni in the activity series. Hence, it is able to displace Ni from its salt  [tex]NiCl_{2}[/tex].

All the other reactions are not feasible based on the activity series.

The link for the activity series I referred is given below:

https://studylib.net/doc/9281468/topic-9.1-activity-series

Answer:answers are in the explanation

Explanation:

(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.

pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.

(b). Equation of reaction;

HBr + KOH ---------> KBr + H2O

One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O

Calculating the mmol, we have;

mmol KOH = 28.0 ml × 0.50 M

mmol KOH= 14 mmol

mmol of HBr= 56 ml × 0.25M

mmol of HBr= 14 mmol

Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.

The pH here is greater than 7

(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL

= 0.10 M

Ka=Kw/kb

10^-14/ 1.8× 10^-5

Ka= 5.56 ×10^-10

Therefore, ka= x^2 / 0.20

5.56e-10 = x^2/0.20

x= (0.20 × 5.56e-10)^2

x= 1.05 × 10^-5

pH = -log [H+]

pH= - log[1.05 × 10^-5]

pH = 4.98

Acidic(less than 7)

(c). 0.5 × 20/40

= 0.25 M

Ka= Kw/kb

kb= 10^-14/1.8× 10^-5

Kb = 5.56×10^-10

x= (5.56×10^-10 × 0.5)^2

x= 1.667×10^-5 M

pH will be basic

Final answer:

In titrations, the pH at the equivalence point depends on the type of titration. For part b, the pH after adding 28.0 mL of KOH can be calculated by determining the moles of HBr and KOH and using the stoichiometry of the reaction. Similarly, for part c and part d, we can calculate the pH by determining the moles of the reactants and using stoichiometry.

Explanation:

For the first part of the question, the pH at the equivalence point depends on the type of titration. In a titration between a weak acid and a strong base, the pH at the equivalence point will be greater than 7. In a titration between a strong acid and a strong base, the pH at the equivalence point will be equal to 7. In a titration between a weak base and a strong acid, the pH at the equivalence point will be less than 7.

For part b, we have a titration between a strong acid (HBr) and a strong base (KOH). The initial volume of HBr is 56.0 mL and the volume of KOH added is 28.0 mL. To calculate the pH after the addition of 28.0 mL of KOH, we need to determine the moles of HBr and KOH and use the stoichiometry of the reaction to find the concentration of the resulting solution, which will give us the pH.

For part c, we have a titration between a weak base (NH3) and a strong acid (HNO3). The initial volume of NH3 is 50.0 mL and the volume of HNO3 added is 50.0 mL. Similarly to part b, we need to determine the moles of NH3 and HNO3 and use the stoichiometry of the reaction to find the concentration of the resulting solution, which will give us the pH.

For part d, we have a titration between a weak acid (CH3COOH) and a strong base (NaOH). The initial volume of CH3COOH is 30.0 mL and the volume of NaOH added is also 30.0 mL. Again, we need to determine the moles of CH3COOH and NaOH and use the stoichiometry of the reaction to find the concentration of the resulting solution, which will give us the pH at this point.

Answer: C) 0.020 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

[tex]Molality=\frac{n\times 1000}{W_s}[/tex]

where,

n = moles of solute  

[tex]W_s[/tex] = weight of solvent in g  

Mole fraction of [tex]CO_2[/tex] is = [tex]3.6\times 10^{-4}[/tex] i.e.[tex]3.6\times 10^{-4}[/tex]  moles of [tex]CO_2[/tex] is present in 1 mole of solution.

Moles of solute [tex](CO_2)[/tex] = [tex]3.6\times 10^{-4}[/tex]

moles of solvent (water) = 1 - [tex]3.6\times 10^{-4}[/tex] = 0.99

weight of solvent =[tex]moles\times {\text {Molar mass}}=0.99\times 18=17.82g[/tex]

Molality =[tex]\frac{3.6\times 10^{-4}\times 1000}{17.82g}=0.020[/tex]

Thus  approximate molality of [tex]CO_2[/tex] in this solution is 0.020 m

Answer:C

Explanation:

A process is called spontaneous if the process takes place on its own without the intervention of external factors.

The spontaneous processes are generally quick with observable rates of reaction.But processes can be spontaneous even with negligible rates of reaction.

Spontaneity of reactions depend on temperature.

This is because spontaneity is measured by gibbs energy or enthalpy.

Both of these measures are dependent on temperature.

So,temperature affects spontaneity.

Answer:

1 billion molecules O₂

Explanation:

From my research, a human red blood cell contains approximately 270 million hemoglobin molecules.    

A hemoglobin molecule contains four heme groups, each of which has an iron ion forming a coordination complex that carries every dioxygen molecule. Therefore for each hemoglobin molecule, we will have 4 dioxygen molecules. The heme groups are responsible for the transport of every dioxygen and other diatomic gases.                    

Hence, the number of O₂ molecules in a red blood cell saturated with 100% will be:                

[tex] \frac{270 \cdot10^{6} hemoglobine molecules}{1 red blood cell} \cdot \frac{4 heme group}{1 hemoglobine molecule} \cdot \frac{1 O_{2} molecules}{1 heme group} = 1 \cdot 10^{9} O_{2} \frac{molecules}{red blood cell} [/tex]

So, the correct answer is 1 billion of O₂ molecules.  

Have a nice day!

Answer:

These properties are basically the inverse of each other.

Explanation:

        Ionization enthalpy, is the energy required to remove an electron from                        an atom.

Ionization energy is the reverse of electronegativity. Ionization energy tells you how easily or how attracted you are to your own electrons so how hard is for somebody to steal them so it's about how much you hold on to your electrons.

Electronegativity is a chemical property and it's about how much can I steal somebody else's electrons. They are like two sides of the same coin. Electronegativity is affected by both its atomic number (number of protons in the core of an atom) and how far the valence electrons (the outermost electrons) are from the core.

a. [tex]\[ n \approx 0.0711 \, \text{moles} \][/tex]

b. the percentage by mass of magnesium carbonate in the mixture is [tex]\(100\%\)[/tex].

a) To find the total number of moles of carbon dioxide formed, we first need to calculate the number of moles of carbon dioxide using the ideal gas law. Then, we can use stoichiometry to relate the moles of carbon dioxide to the moles of magnesium carbonate reacted.

Given:

- Mass of mixture = 6.53 g

- Volume of carbon dioxide = 1.73 L

- Temperature = 26 °C = 26 + 273.15 K = 299.15 K

- Pressure = 745 torr

First, let's calculate the number of moles of carbon dioxide using the ideal gas law:

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

Where:

- P is the pressure in atm (convert 745 torr to atm),

- V is the volume in liters,

- n is the number of moles,

- R is the gas constant (0.0821 atm L/mol K),

- T is the temperature in Kelvin.

[tex]\[ P = \frac{745 \, \text{torr}}{760 \, \text{torr/atm}} = 0.980 \, \text{atm} \][/tex]

[tex]\[ n = \frac{PV}{RT} = \frac{(0.980 \, \text{atm})(1.73 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K})(299.15 \, \text{K})} \][/tex]

[tex]\[ n \approx 0.0711 \, \text{moles} \][/tex]

b) Now, let's use stoichiometry to relate the moles of carbon dioxide formed to the moles of magnesium carbonate reacted. From the balanced chemical equation:

[tex]\[ \text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]

We see that 1 mole of magnesium carbonate [tex](\( \text{MgCO}_3 \))[/tex] reacts to produce 1 mole of carbon dioxide [tex](\( \text{CO}_2 \))[/tex].

So, the moles of magnesium carbonate reacted is also [tex]\(0.0711 \, \text{moles}\)[/tex].

Now, let's use the mass of magnesium carbonate and the moles reacted to find the percentage:

[tex]\[ \text{mass of MgCO}_3 = 6.53 \, \text{g} \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = \frac{\text{mass of MgCO}_3}{\text{total mass of mixture}} \times 100\% \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = \frac{6.53 \, \text{g}}{\text{total mass of mixture}} \times 100\% \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = \frac{6.53 \, \text{g}}{6.53 \, \text{g}} \times 100\% \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = 100\% \][/tex]

Therefore, the percentage by mass of magnesium carbonate in the mixture is [tex]\(100\%\)[/tex].

(a)The total number of moles of carbon dioxide that forms is [tex]\( 0.0691 \)[/tex] moles. (b) The percentage by mass of magnesium carbonate in the mixture is [tex]\( 30.9\% \)[/tex].

To solve this problem, we need to use the given data to find the total number of moles of carbon dioxide produced and then use stoichiometry to determine the percentage by mass of magnesium carbonate in the mixture.

Part (a): Calculate the total number of moles of carbon dioxide

First, we use the ideal gas law to find the number of moles of carbon dioxide gas produced. The ideal gas law is given by:

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

Where:

- P is the pressure in atmospheres (atm)

- V is the volume in liters (L)

- n is the number of moles

- R is the ideal gas constant [tex](0.0821 LatmK\(^{-1}\)mol\(^{-1}\))[/tex]

- T is the temperature in Kelvin (K)

We need to convert the given pressure from torr to atm and the temperature from Celsius to Kelvin.

[tex]\[ P = 745 \, \text{torr} \times \frac{1 \, \text{atm}}{760 \, \text{torr}} = 0.980 \, \text{atm} \][/tex]

[tex]\[ T = 26^\circ \text{C} + 273 = 299 \, \text{K} \][/tex]

Given [tex]\( V = 1.73 \, \text{L} \)[/tex], we can solve for n:

[tex]\[ n = \frac{PV}{RT} \][/tex]

[tex]\[ n = \frac{(0.980 \, \text{atm})(1.73 \, \text{L})}{(0.0821 \, \text{LatmK}^{-1}\text{mol}^{-1})(299 \, \text{K})} \][/tex]

[tex]\[ n = \frac{1.6954 \, \text{atmL}}{24.5479 \, \text{LatmK}^{-1}\text{mol}^{-1}} \][/tex]

[tex]\[ n = 0.0691 \, \text{mol} \][/tex]

Part (b): Calculate the percentage by mass of magnesium carbonate in the mixture

The reactions are:

[tex]\[ \text{MgCO}_3 (s) + 2\text{HCl} (aq) \rightarrow \text{MgCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \][/tex]

[tex]\[ \text{CaCO}_3 (s) + 2\text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \][/tex]

Both reactions produce [tex]\( \text{CO}_2 \)[/tex] gas in a 1:1 molar ratio with their respective carbonates.

Let x be the mass of [tex]\( \text{MgCO}_3 \)[/tex] and y be the mass of [tex]\( \text{CaCO}_3 \)[/tex]. We know:

[tex]\[ x + y = 6.53 \, \text{g} \][/tex]

The moles of [tex]\( \text{CO}_2 \)[/tex] produced from each carbonate are:

[tex]\[ \frac{x}{84.31} \, \text{mol} \, (\text{for} \, \text{MgCO}_3) \][/tex]

[tex]\[ \frac{y}{100.09} \, \text{mol} \, (\text{for} \, \text{CaCO}_3) \][/tex]

The total moles of [tex]\( \text{CO}_2 \)[/tex] is the sum of the moles produced by each carbonate:

[tex]\[ \frac{x}{84.31} + \frac{y}{100.09} = 0.0691 \][/tex]

We have two equations:

1. [tex]\( x + y = 6.53 \)[/tex]

2. [tex]\( \frac{x}{84.31} + \frac{y}{100.09} = 0.0691 \)[/tex]

Solve these equations simultaneously. First, solve equation 1 for y:

[tex]\[ y = 6.53 - x \][/tex]

Substitute into equation 2:

[tex]\[ \frac{x}{84.31} + \frac{6.53 - x}{100.09} = 0.0691 \][/tex]

Multiply through by [tex]\( 84.31 \times 100.09 \)[/tex] to clear the denominators:

[tex]\[ 100.09x + 84.31(6.53 - x) = 0.0691 \times 84.31 \times 100.09 \][/tex]

[tex]\[ 100.09x + 84.31 \times 6.53 - 84.31x = 582.5 \][/tex]

Combine like terms:

[tex]\[ 15.78x + 550.56 = 582.5 \][/tex]

Solve for [tex]\( x \)[/tex]:

[tex]\[ 15.78x = 31.94 \][/tex]

[tex]\[ x = 2.02 \, \text{g} \][/tex]

Now, calculate [tex]\( y \)[/tex]:

[tex]\[ y = 6.53 - 2.02 = 4.51 \, \text{g} \][/tex]

Finally, calculate the percentage by mass of magnesium carbonate in the mixture:

[tex]\[ \text{Percentage of MgCO}_3 = \left( \frac{2.02 \, \text{g}}{6.53 \, \text{g}} \right) \times 100\% \][/tex]

[tex]\[ \text{Percentage of MgCO}_3 = 30.9\% \][/tex]

Explanation:

Endothermic animals are also known as warm-blooded, they have the capacity to regulate their body temperature independent of the environment. They have mechanisms to compensate if heat loss exceeds heat generation (shivers) Or if heat generation exceeds the heat loss (panting, sweating).

On the other hand, ectothermal animals are known as cold blooded organisms and depend on external sources, like sunlight, to regulate their body temperature, reptiles are ectothermals.

To determine if the animal of interest is endo or ectothermal you’ll have to consider that is a reptile, you’ll also observe that it consumes less food and finally it’ll have more difficulties to adapt to sudden temperature changes.

I hope you find this information useful and interesting! Good luck!

To determine if a tropical reptile is an endotherm or an ectotherm, observe its behavior in response to environmental temperature changes and test its metabolic rate and body temperature in various conditions.

To determine whether a large tropical reptile with a high and relatively stable body temperature is an endotherm or an ectotherm, one should observe the animal's behavior and response to changes in environmental temperature. If the reptile is an endotherm, it would generate its own heat through metabolic processes and maintain a stable body temperature regardless of the environment.

Endotherms demonstrate behaviors like shivering to generate heat when cold. Ectotherms, on the other hand, rely on the environment to regulate their body temperature, and this can be observed if the animal basks in the sun to warm up or seeks shade to cool down.

Most reptiles are known to be ectotherms and can exhibit behaviors such as brumation in response to cold temperatures. This state is similar to hibernation but doesn't rely on fat reserves. Therefore, if the reptile is less active in colder conditions and depends on external heat sources like sunlight, it is likely an ectotherm. However, if the reptile maintains an active metabolism and body temperature in colder environments without relying on external heat sources, it may be an endotherm.

Specific tests could include monitoring the reptile's body temperature in controlled environments with varying temperatures and measuring the metabolic rate under these conditions to see if there is internal heat production.

Answer : The molar mass of catalase is, [tex]2.40\times 10^5g/mol[/tex]

Explanation :

Formula used :

[tex]\pi =CRT\\\\\pi=\frac{w}{M\times V}RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure  = 0.745 torr = 0.000980 atm   (1 atm = 760 torr)

C = concentration

R = solution constant  = 0.0821 L.atm/mol.K

T = temperature  = [tex]27^oC=273+27=300K[/tex]

w = mass of catalase = 10.03 g

M = molar mass of catalase = ?

V = volume of solution  = 1.05 L

Now put all the given values in the above formula, we get:

[tex]0.000980atm=\frac{10.03g}{M\times 1.05L}\times (0.0821L.atm/mole.K)\times (300K)[/tex]

[tex]M=2.40\times 10^5g/mol[/tex]

Therefore, the molar mass of catalase is, [tex]2.40\times 10^5g/mol[/tex]

The molar mass of catalase is 326.99 g/mol.

To calculate the molar mass of catalase, we can use the formula for osmotic pressure, II = MRT, where M is the molar concentration of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the formula to solve for M, we get M = II / RT. Plugging in the given values, II = 0.745 torr, R = 0.08206 L atm/mol K, and T = 27°C = 300 K, we can solve for M. M = (0.745 torr) / (0.08206 L atm/mol K) / 300 K = 0.0306 mol/L. Since the solution has a volume of 1.05 L and contains 10.03 g of catalase, we can use the formula M = mass / (volume * molar mass) to solve for the molar mass. Rearranging the formula to solve for molar mass, we get molar mass = mass / (volume * M). Plugging in the given values, molar mass = 10.03 g / (1.05 L * 0.0306 mol/L) = 326.99 g/mol.

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Answer:

B

Explanation:

In counting the electron domains around the central atom in VSEPR theory, a Core level electron pair is not included.

Core level electron pair are the electrons other than the electrons in the valency shell of an atom. They in the proximity of the nucleus. They do not participate in the bonding.

VSEPR is the abbreviation for Valence shell Electron Pair repulsion theory. VESPR is a model for predicting molecular geometries based on the reduction in the electrostatic repulsion of the valence electrons of a molecule around a central atom.

Answer:

The correct answer is a A molecule that brings about a cellular response to a signal.

Explanation:

Effector molecule is a small molecule that act as ligand to generate various cellular response by binding with target proteins(receptors).

  Effector molecules can regulate catalytic activity of enzymes,gene expression and various signaling processes that are very much important for proper functioning of the cell that the effect molecule target.

 Examples Allosteric effectors which can be either modulators or inhibitors depending on the nature of the effector molecule.

Answer:

what are your statements?

Answer:

b) O2, because it has weaker intermolecular forces

Explanation:

The preassure is produced by the collisions of the gas molecules with the walls of its container.

When the intermolecular forces between the gas molecules increase, those molecules start to "slow down" by effect of the interactions. The collisions decrease in frequency and intensity producing a smaller preassure in the container.

Both O2 and Cl2 are non-polar gases and the only intermolecular forces they have are the London ones. Given that the O2 molecules are smaller than the Cl2, the last ones attract each other with more strengh.

Being all that said, the container with the oxygen is expected to have a higher preassure.

Oxygen molecule exert a higer pressure because it has weaker intermolecular forces.

The pressure is produced by the collisions of the gas molecules with the walls of its container. If the intermolecular forces between the gas molecules increase, those molecules start to slow down which leads to lowering of pressure on the wall of the container.

So we can conclude that decreasing intermolecular force, increase the pressure on the wall of the container.

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Answer:

1)50.007 grams of an isotope will left after 10 years.

1)25.007 grams of an isotope will left after 20 years.

3) 23 half-lives will occur in 40 years.

Explanation:

Formula used :

[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]

where,

[tex]N_o[/tex] = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope

[tex]\lambda[/tex] = rate constant

We have:

[tex][N_o]=100 g[/tex]

[tex]t_{1/2}=10 years[/tex]

[tex]\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{10 year}=0.0693 year^{-1}[/tex]

1) mass of isotope left after 10 years:

t = 10 years

[tex]N=N_o\times e^{-\lambda t}[/tex]

[tex]N=100g\times e^{-0.0693 year^{-1}\times 10}=50.007 g[/tex]

2) mass of isotope left after 20 years:

t = 20 years

[tex]N=N_o\times e^{-\lambda t}[/tex]

[tex]N=100g\times e^{-0.0693 year^{-1}\times 20}=25.007 g[/tex]

3) Half-lives will occur in 40 years

Mass of isotope left after 40 years:

t = 40 years

[tex]N=N_o\times e^{-\lambda t}[/tex]

[tex]N=100g\times e^{-0.0693 year^{-1}\times 40}=6.2537 g[/tex]

number of half lives = n

[tex]N=\frac{N_o}{2^n}[/tex]

[tex]6.2537 g=\frac{100 g}{2^n}[/tex]

[tex]n\ln 2=\frac{100 g}{6.2537 g}[/tex]

n = 23

23 half-lives will occur in 40 years.

Answer:

AS we move from bottom to top on periodic table shielding decreased.

Explanation:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction.

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As we move from bottom to top the energy level decreased because of decreased in electron thus shielding decreased and atomic size also decreased.

Answer: Glucose and oxygen

Explanation:

The basic products of photosynthesis are glucose and oxygen. Photosynthesis is a process in which green plants produce there own food from carbon dioxide (CO2) and water (H2O) and in the presence of energy from the sun to produce glucose(food) and oxygen

After series of reaction, the overall equation for photosynthesis is

6CO2 + 6H20 + (sunlight ) → C6H12O6 + 6O2

Carbon dioxide + water + energy from sunlight to produces glucose and oxygen.

Answer is glucose and oxygen

Answer:

1. False

2. False

3. True

Explanation:

Identify all statements are true

1. zeroth order rate constants have units of inverse seconds. FALSE.

The rate law for a zeroth order rate reaction has the following form:

r = k

where,

r is the rate of the reaction

k is the rate constant

k has the same units than r, that is, M . s⁻¹.

2. first order reactions have half lives that are dependent on time . FALSE.

Half-life (t1/2) of a first order reaction can be calculated using the following expression.

[tex]t_{1/2}=\frac{ln2}{k}[/tex]

As we can see, half-life does not depend on time.

3. catalysts cannot appear in the rate law for a reaction. TRUE.

The rate law has the following form:

r = k . [A]ᵃ . [B]ᵇ

where,

[A] and [B] are the concentrations of the reactants

a and b are the reaction orders

Catalysts do not appear in the rate law.

Final answer:

To determine the time taken for the decay rate to drop from 15 to 4 disintegrations per minute using a half-life of 5730 years, we calculate that it would take two half-lives, or 11460 years, for this reduction.

Explanation:

To calculate the time it would take for a piece of wood with an initial decay rate of 15 disintegrations per minute to decrease to 4 disintegrations per minute, we can use the concept of half-life, which for carbon-14 is 5730 years. By applying the decay constant and the exponential decay formula, we can find the time that corresponds to the rate of decay reaching 4 disintegrations per minute.

The relationship we use is N = N0e-λt, where N is the final number of disintegrations per minute, N0 is the initial number of disintegrations per minute, λ is the decay constant (which can be calculated using λ = 0.693 / 5730 years), and t is the time in years.

Using the concept of half-lives, we can determine that if after t years the disintegration rate is a quarter of its original rate (15 to 4 dpm), the wood must have gone through two half-lives, because each half-life reduces the rate of disintegration by half. Therefore, t would be equal to 2 × 5730 years = 11460 years.

Answer:

[tex]\large \boxed {\text{24.3 g Cl}_{2}; 82.3 \%}[/tex]

Explanation:

We are given the masses of two reactants and asked to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

MM:          86.94    36.46                 70.91

                MnO₂ + 4HCl ⟶ MnCl₂ + Cl₂ + 2H₂O

Mass/g:     86.0      50.0                  20.00

2. Calculate the moles of each reactant

[tex]\text{Moles of MnO}_{2} = \text{86.0 g MnO}_{2} \times \dfrac{\text{1 mol MnO}_{2}}{\text{86.94 g MnO}_{2}} = \text{0.9892 mol MnO}_{2}\\\\\text{Moles of HCl} = \text{50.0 g HCl} \times \dfrac{\text{1mol HCl }}{\text{36.46 g HCl }} = \text{1.371 mol HCl}[/tex]

3. Calculate the moles of Cl₂ formed from each reactant

From MnO₂:

[tex]\text{Moles of Cl$_{2}$} = \text{0.9892 mol MnO}_{2} \times \dfrac{\text{1 mol Cl$_{2}$}}{\text{1 mol MnO}_{2}} = \text{0.9892 mol Cl}_{2}[/tex]

From HCl:

[tex]\text{Moles of Cl$_{2}$} = \text{1.371 mol HCl} \times \dfrac{\text{1 mol Cl$_{2}$}}{\text{4 mol HCl}} = \text{0.3428 mol Cl}_{2}[/tex]

4. Identify the limiting reactant

The limiting reactant is HCl, because it forms fewer moles of Cl₂.

5. Calculate the theoretical yield of Cl₂

[tex]\text{ Mass of Cl$_{2}$} = \text{0.3428 mol Cl$_{2}$} \times \dfrac{\text{70.91 g Cl$_{2}$}}{\text{1 mol Cl$_{2}$}} = \textbf{24.3 g Cl}_\mathbf{{2}}\\\\\text{The theoretical yield is $\large \boxed{\textbf{24.3 g Cl}_\mathbf{{2}}}$}[/tex]

6. Calculate the percentage yield of Cl₂

[tex]\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \%= \dfrac{\text{20.0 g}}{\text{24.3 g}} \times 100 \, \% = 82.3 \, \%\\\text{The percentage yield is $\large \boxed{\mathbf{82.3 \, \%}}$}[/tex]

Answer:

Sp^3d

Explanation:

Sp^3d hybridization invovles ther combination of 1s, 3p and 1d orbital  to result in the production of sp3d orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions. An important and relatively common type of this hybridization is found in Clf3.

In ClF3  (Chlorine Trifluoride), The central atom Cl needs three unpaired electrons to bond with three F-atoms. ClF3 consist of 3 bond-pairs and 2 lone-pairs. Here, one of the paired electrons of Cl in the 3p subshell remains as a lone pair or unpaired. During hybridisation, one 3s, three 3p and one of the 3d orbitals participate in the process which leads to the formation of five sp3d hybrid orbitals. Here two hybrid orbitals will contain a pair of electrons and three hybrids will contain unpaired electrons which will again overlap with the 2p orbital of F to form single bonds.

Answer:

E) Scandium (Sc)

Hope this helps!

Here 'n' represents the principal quantum number. The element which has n=4 and has three electrons in its valence p orbital is scandium (Sc). The electronic configuration of 'Sc' is [Ar] 3d¹ 4s². The correct option is E.

The quantum number which represents the main energy level or shell in which the electron is present. It also determines the average distance of the orbital or electron from the nucleus. It is denoted by letter 'n' and can have any whole number values like 1, 2, 3, 4, etc. These values represent different energy levels.

The atomic number of the element scandium is 21 and its electronic configuration is [Ar] 3d¹ 4s². The total number of valence electrons present here is 3 and the energy level is 4.

The element scandium is a transition element and it has the highest melting point and low density.

Thus the correct option is E.

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