An isothermal chromatogram at 90 °C shows an elution order of 1-pentanol followed by ethylene diamine followed by diethylene glycol. But in order to reduce the overall time of the experiment, a temperature programmed chromatogram is tried, beginning at 60 °C and increasing to 190 °C over 5 minutes then holding at 190 °C for 4 minutes. Predict the elution order of the compounds using the temperature programmed conditions.

We calculated that after the 5.00 g of calcium oxide reacts with water to produce calcium hydroxide, the rest of the water left is 3.23g.

To solve this question, we first need to understand the stoichiometry of the reaction. The balanced equation clearly illustrates that one mole of calcium oxide (CaO) reacts with one mole of water (H2O) to form calcium hydroxide (Ca(OH)2). Therefore, this reaction is a 1:1 ratio.

Next, we need to convert the grams to moles. Since the molar mass of CaO is approximately 56.08 g/mol and H2O is 18.015 g/mol, we can calculate that 5.00 g of CaO is around 0.089 mol and 4.83 g of H2O is approximately 0.268 mol.

Considering the stoichiometry of the reaction, it is clear that not all of the water will react because it is present in excess. Only an amount equivalent to the moles of CaO will, so 0.089 mol of H2O will react. To convert this back to grams, simply multiply the moles of water reacted by the molar mass of water. This is approximately 1.60g.

Then, subtract the amount of water used in the reaction from the original amount to get the amount of water left. Hence, 4.83 g - 1.60 g = 3.23 g of water remain after the reaction is complete.

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Answer:

-Hybridization does not account for observed bond angles in molecules.

-Single bonds are always pi bonds.

-The length of a bond is determined by where the energy of the system is at its lowest point                                                                

Explanation:

-The very first statement is incorrect because it does account for different bong angles as the hybrid orbitals are responsible for contributing for bond angles in a way that more the hybrid orbitals present the lesser the angles it forms.

-The second statement is incorrect because single bonds are considered as sigma bonds and not a pi bond.

-The third statement is incorrect because hybridization is responsible for deciding the bond length.

The incorrect statements about bonding and hybridization are that single bonds are always pi bonds and pi bonds are always between unhybridized p orbitals.

The incorrect statements about bonding and hybridization are:

Hybridization does not account for observed bond angles in molecules, so this statement is correct. The length of a bond is determined by where the energy of the system is at its lowest point, so this statement is also correct. Multiple bonds can have a combination of sigma and pi bonds, so this statement is correct as well.

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Answer:

1.496x 10^24photons

Explanation:

wavelength λ= 680 X10^-9 nm

h = planks constant - 6.636*10 ^-34js

c- speed of light - 3.0x 10^8 m/s

I mole of  Energy of NADP+ = 219Kj/mol

2 moles of  Energy of NADP+ = 2x 219= 438kj/mol = 438x10^3j

/mol

E= nhc/λ

438x 10^3j/mol -= n x (6.636*10 ^-34 x  3x10^8) / 680*10^-9

n=438x10^3j x 680x 10^-9/ (6.636*10 ^-34 x 3.0x10^8

1.496x 10^24photons

To make the reduction of 2 moles of NADP+ favorable for light absorbed at 680.000 nm, you would need a minimum of approximately [tex]6.35 x 10^{15[/tex] photons, which are particles of light.

The energy required for the reduction of 2 moles of NADP+ is given as 2 moles x 219 kJ/mol = 438 kJ. To calculate the number of photons needed, we can use the formula E = nhf, where E is the energy required, h is Planck's constant (6.626 x [tex]10^{-34[/tex] J·s), f is the frequency of light, and n is the number of photons. First, we need to convert the given wavelength to frequency using the speed of light (c =3 x[tex]10^8[/tex] m/s):

λ = 680.000 nm = 680.000 x [tex]10^{-9[/tex] m

f = c / λ = (3 x [tex]10^8[/tex] m/s) / (680.000 x [tex]10^{-9[/tex] m) ≈ 4.41 x [tex]10^{14[/tex] Hz

Now, we can calculate the number of photons (n) using the energy formula:

E = nhf

438,000 J = n(6.626 x [tex]10^{-34[/tex] J·s)(4.41 x [tex]10^{14[/tex] Hz)

Solving for n:

n ≈ 6.35 x [tex]10^{15[/tex] photons

So, approximately 6.35 x [tex]10^15[/tex]photons are needed to make the reduction of 2 moles of NADP+ favorable for light absorbed at 680.000 nm.

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To convert atoms to moles, divide the number of atoms by Avogadro's number of 6.02 x 10²³ atoms per mole.

Your friend is partially correct. In order to convert atoms to moles, you use the constant known as Avogadro's number (6.02 x 10²³ atoms per mole). However, it's important to note that you need to divide the number of atoms by Avogadro's number, not multiply it. Let's give an example:

Example: If we have 2.56 x 10²⁴ atoms of Uranium, we'd use Avogadro's number to convert this to moles like so: (2.56 x 10²⁴ atoms) / (6.02 x 10²³ atoms/mol) = 4.25 moles of Uranium

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No, your friend is not correct. To convert atoms to moles, divide the number of atoms by Avogadro's number, 6.022 × 10²³. This is because 1 mole of any substance contains 6.022 × 10²³ atoms.

To convert the number of atoms to moles, you should divide the number of atoms by Avogadro's number, which is 6.022 × 10²³. This relationship is based on the fact that 1 mole of any substance contains exactly 6.022 × 10²³ atoms, a constant known as Avogadro's number.

Step-by-Step Explanation:

For example, if you have 1.2044 × 10²⁴ atoms of hydrogen:

Complete Question: -

I want to convert atoms to moles. My friend tells my to multiply the number of atoms by 6.02 x 10²³. Is my friend correct?

Answer:

The Answer Is A. W: Fahrenheit Y: Kelvin X: Celsiuis

Answer:

a

Explanation:

Answer:

I can provide a proper answer since there are no bonds specified.

Explanation:

Answer:

I THINK mechanical energy

Answer:

The additional information required to solve this problem is the initial volume.

the final pressure P₂ of the gas is 1.108 atm

Explanation:

Given that :

A sample of gas at initial temperature [tex]T_1 = 12.0^0 \ C[/tex] = (12+273)K = 285 K

Pressure (P₁) = 1.06 atm

Initial Volume (V₁) = unknown ???

Final Volume (V₂) = 2.30 L

final temperature [tex]T_2 = 24.9^0 \ C[/tex]  = (24.9 +273)K = 297.9 K

Find the final Pressure (P₂)

The relation between: Pressure, Volume and Temperature can be gotten from the ideal gas equation :

PV = nRT

The Ideal Gas Equation is also reduced to the General Gas Law or the combined Gas Law by assuming that n= 1 .

From ; PV = nRT

[tex]\frac{PV}{T} = R \ \ ( constant) \ if \ n=1[/tex]

∴ [tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = \frac{P_3V_3}{T_3}...= \frac{P_nV_n}{T_n} \ \ \ ( n \ constant)[/tex]

The additional information required to solve this problem is the initial volume.

This expression is a combination of Boyle's Law and Charles Law. From the combined Gas Law , it can be deduced that at constant volume, the pressure of a given mass(mole) of gas varies directly with absolute temperature.

∴ [tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]  if n & Volume (V) are constant .

[tex]P_2 = \frac{1.06*297.9}{285}[/tex]

P₂ = 1.108 atm

Thus, the final pressure P₂ of the gas is 1.108 atm

Answer:

x= 138.24 g

Explanation:

We use the avogradro's number

6.023 x 10^23 molecules -> 1 mol C2H8

26.02 x 10^23 molecules -> x

x= (26.02 x 10^23 molecules  * 1 mol C2H8 )/6.023 x 10^23 molecules

x= 4.32 mol C2H8

1 mol C2H8     -> 32 g

4.32 mol C2H8 -> x

x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8

x= 138.24 g

The correct answer is 156.69 * 10^46 grams.

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Approximately 0.237 milligrams of cesium-137 would remain after 60 years.

The amount of a radioactive substance that remains after a certain amount of time can be calculated using the decay constant. For cesium-137, the decay constant is 0.0871 per year. To determine the amount remaining after 60 years, we can use the formula:
Amount remaining = initial amount * e^(-decay constant * time)
Substituting the values, we get:
Amount remaining = 20mg * e^(-0.0871 * 60) = 20mg * e^(-5.226) ≈ 0.237mg. Therefore, approximately 0.237 milligrams of cesium-137 would remain after 60 years.

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Answer:

Exper

des

com

Explanation:

Final answer:

Experimental investigations focus on manipulating one variable and controlling others to determine effects, while descriptive investigations observe natural occurrences without manipulation. Field experiments modify a variable in a natural environment with some control over extraneous factors.

Explanation:

In scientific investigations, it is essential to understand the roles of different types of variables and controls. Experimental investigations allow for the control of variables to ensure that only one variable is manipulated, which is the independent variable. This isolation helps in distinguishing the direct effects of the manipulation on the dependent variable, which is being measured and recorded.

Field investigations provide an opportunity to study phenomena in a natural setting, where controlling all extraneous variables is not always feasible. However, field experiments can also be conducted where one independent variable is intentionally altered while attempting to control extraneous factors, thus achieving a balance between external and internal validity.

Lastly, there are observational studies or descriptive investigations which do not manipulate variables, but rather observe and record variables as they naturally occur. These are typically less expensive, less time-consuming, and can encompass a wide range of variables, although they often lack the control of experimental studies.

Logical steps to do the investigation involve identifying the independent, dependent, and controlled variables, establishing controls and a control group if applicable, and following an experimental procedure that ensures repeatability and reliability of the results.

Answer:

Time is something I constantly run out of-

Answer:

The concept of time is self-evident. An hour consists of a certain number of minutes, a day of hours and a year of days. ... Time is represented through change, such as the circular motion of the moon around Earth. The passing of time is indeed closely connected to the concept of space

Answer:

=> 572.83 K (299.83°C).

=> 95.86 m^2.

Explanation:

Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.

Hence;

Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.

15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).

If we solve for x, we will get the value of x to be;

x = 572.83 K (2.99.83°C).

x is the temperature of the exit gas that is 572.83 K(299.83°C).

(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).

∆T = 93.87/ln 1.38.

∆T = 291.521K.

Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.

heat-transfer area = 1931146.394/69.1 × 291.521.

heat-transfer area= 95.86 m^2.

Answer: Amount of Gallium = 0.127g

Explanation:

Electrolysis equation is:

Ga3+   +   3e-    ------> Ga

 To calculate the charge

t = 40.0 min = 40.0 x 60 s = 2400 s

time, t = 2400s

Q = I*t =  

= 0.22A x 2400s

= 528 C

1 mol of Ga requires 3 mol of electron

1 mol of electron = 1 Faraday =96485 C

So,1 mol of Ga requires  96485x 3= 289455 C

mol of Gallium = 528/289455 = 0.00182 mol

Molar mass of Ga = 69.72 g/mol

mass of Ga = number of moles x  molar mass

= 0.00182mol * 69.72

g/mol

= 0.127g

or you can use this direct formula

m=(current*time/Faraday's)*(molar mass/no of electrons transferred)

keeping in mind   Ga3+ + 3e- → Ga

n=3

m=(It/F)*(mew/n)

m =(0.22 x 2400/96485) x (69.72/3)

m=0.127 g

Answer:

Explanation:

Given that:

the temperature [tex]T_1[/tex] = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

[tex]\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}[/tex]

where; B = - [tex]152.5 \ cm^3 /mol[/tex]   C = -5800 [tex]cm^6/mol^2[/tex]

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

[tex]\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]

[tex]4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]

Multiplying through with V² ; we have

[tex]4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0[/tex]

[tex]4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0[/tex]

V = 2250.06  cm³ mol⁻¹

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

Z = [tex]\frac{1800*2250.06}{8.314*10^3*523.15}[/tex]

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

[tex]T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345[/tex]

[tex]T__{\gamma}} = \frac{T}{T_c}[/tex]

[tex]T__{\gamma}} = \frac{523.15}{647.1}[/tex]

[tex]T__{\gamma}} = 0.808[/tex]

[tex]P__{\gamma}} = \frac{P}{P_c}[/tex]

[tex]P__{\gamma}} = \frac{1800}{22055}[/tex]

[tex]P__{\gamma}} = 0.0816[/tex]

[tex]B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}[/tex]

[tex]B_o = 0.083 - \frac{0.422}{0.808^{1.6}}[/tex]

[tex]B_o = 0.51[/tex]

[tex]B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}[/tex]

[tex]B_1 = -0.282[/tex]

The compressibility is calculated as:

[tex]Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}[/tex]

[tex]Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}[/tex]

Z = 0.9386

[tex]V= \frac{ZRT}{P}[/tex]

[tex]V= \frac{0.9386*8.314*10^3*523.15}{1800}[/tex]

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At [tex]T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa[/tex]

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

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

Z = [tex]\frac{1800*10^3 *0.1249}{729.77*523.15}[/tex]

Z = 0.588

Final answer:

To determine Z and V for steam at 250°C and 1800 kPa, we can use the truncated virial equation with the given experimental values of the virial coefficients, or we can use the generalized Pitzer correlation to obtain the B value and then use the truncated virial equation. Alternatively, we can look up the values in the steam tables.

Explanation:

The question asks us to determine Z and V for steam at 250°C and 1800 kPa using three different methods: (a) The truncated virial equation with given experimental values of the virial coefficients, (b) The truncated virial equation with B value obtained using the generalized Pitzer correlation, and (c) The steam tables.

(a) To determine Z and V using the truncated virial equation with B and C values, we substitute the given temperature and pressure into the equation and solve for Z and V.

(b) To determine Z and V using the truncated virial equation with the B value obtained from the generalized Pitzer correlation, we substitute the given temperature and pressure into the equation and solve for Z and V.

(c) To determine Z and V using the steam tables, we look up the values for Z and V at the given temperature and pressure.

Answer:pH = 2.96

Explanation:

C5H5N + HBr --------------> C5H5N+  + Br-

millimoles of pyridine = 80 x 0.3184 =25.472mM

25.472 millimoles of HBr must be added to reach equivalence point.

25.472  = V x 0.5397

V =25.472/0.5397= 47.197 mL HBr

total volume = 80 + 47.197= 127.196 mL

Concentration of [C5H5N+] = no of moles / volume=

25.472/ 127.196= 0.20M

so,

pOH = 1/2 [pKw + pKa + log C]

pKb = 8.77

pOH = 1/2 [14 + 8.77 + log 0.20]

pOH = 11.0355

pH = 14 - 11.0355

pH = 2.96

Final answer:

To calculate the pH at equivalence, we need to determine the concentration of pyridine and its conjugate acid. The pH at equivalence can be calculated using the Henderson-Hasselbalch equation, which relates the pH to the pKa and the concentration of the conjugate acid and base. The pKa value for the pyridinium ion can be determined by subtracting the pKb of pyridine from the pKw.

Explanation:

To calculate the pH at equivalence, we need to determine the concentration of pyridine and its conjugate acid. From the given information, we know that the initial volume of pyridine solution is 80.0 mL and its concentration is 0.3184 M. We also have the concentration of HBr solution, which is 0.5397 M. The reaction between pyridine and HBr is:

C5H5N (aq) + HBr (aq) → C5H5NH+Br- (aq)

This reaction forms the pyridinium ion (C5H5NH+) which is the conjugate acid of pyridine. At equivalence, the moles of pyridine and pyridinium ion are equal. Using the stoichiometry of the reaction, we can calculate the number of moles of pyridine:

Moles of pyridine = Volume of pyridine solution * Concentration of pyridine = 80.0 mL * 0.3184 M = 25.472 moles

Since the reaction is 1:1, the moles of pyridine also correspond to the moles of pyridinium ion. Therefore, the concentration of pyridinium ion is:

Concentration of pyridinium ion = Moles of pyridinium ion / Volume of pyridinium ion solution = 25.472 moles / 80.0 mL = 0.3184 M

Now, we can use the Henderson-Hasselbalch equation to calculate the pH at equivalence:

pH = pKa + log10 ([A-] / [HA])

Given that the pKb of pyridine is 8.77, we can determine the pKa of pyridinium ion:

pKa = 14.00 - pKb = 14.00 - 8.77 = 5.23

Substituting the values into the Henderson-Hasselbalch equation:

pH = 5.23 + log10 (0.3184 / 0.3184) = 5.23 + 0 = 5.23

Therefore, the pH at equivalence is 5.23.

Answer:

See Explaination

Explanation:

We can define an oxidizing agent as a reactant that removes electrons from other reactants during a redox reaction. The oxidizing agent typically takes these electrons for itself, thus gaining electrons and being reduced. An oxidizing agent is thus an electron acceptor.

Please kindly check attachment for further solutions in details.

In this example, the strongest reducing agent is Aluminum (Al) and the strongest oxidizing agent is the dichromate ion (Cr₂O₇²¯). They both involve 6 electrons in their respective half-reactions. Balancing and summing these half-reactions give the net redox reaction.

To answer your experimental lab results query about redox reactions, we first identify our strongest oxidizing agent and reducing agent. An oxidizing agent is a substance that tends to oxidize other substances, meaning it is reduced. Conversely, a reducing agent reduces other substances while being oxidized itself.

Your strongest reducing agent might be represented by this half-reaction: 2Al(s) → 2Al³+ + 6e⁻, and your strongest oxidizing agent might be represented by: Cr₂O₇²¯ + 14H⁺ + 6e⁻ → 2Cr³+ + 7H₂0. Here, aluminum (Al) is losing electrons, so it's oxidized, and dichromate ion (Cr₂O₇²¯) is gaining electrons, so it's reduced.

So, identifying the number of electrons involved: Reducing half-reaction (Aluminum) = 6 electrons, Oxidizing half-reaction (Chromium) = 6 electrons. As the number of electrons in both half-reactions is equal, no multiplication is needed to balance them.

Now, combining these half-reactions to form the net redox reaction gives us: 2Al(s) + Cr₂O₇²¯ + 14H⁺ → 2Al³+ + 2Cr³+ + 7H₂0.

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Answer:

3)They increase precipitation on the windward sides of mountain ranges

4)They force cool, moist air from oceans to rise as they move toward land

5)They decrease precipitation totals on the leeward sides of mountain ranges.

Explanation:

During the day, as the ocean becomes heated up, there is always an increase in humidity of the air that is above the oceans in coastal areas. But Because of the higher heat capacity of water compare to the land,the ocean always remain cooler compare to the land. But because of lower density,the air on the land is replaced by the ocean's cool air as the land's air increases.

Whenever a mountain inland come in contact with cool

air with humidity,the sir get more cool. However,. there result a condensation and precipitation of the water that is present in the air at the part of the mountain.

Answer:

3.They increase precipitation on the windward sides of mountain ranges.

4.They force cool, moist air from oceans to rise as they move toward land.

5.They decrease precipitation totals on the leeward sides of mountain ranges.

Explanation:

The temperature on mountains usually become colder with a corresponding higher altitude. The Mountains also tend to have more precipitation than areas without them.

This is because in the day the ocean becomes heated up. This leads to an increase in humidity of the air that is above the oceans in coastal areas. Due to the higher heat capacity of water compared to land ,the ocean always remains cooler when compared to the land. The lower density also allows the air on the land to be replaced by the ocean's cool air as the land's air increases.

They force cool, moist air from oceans to rise as they move toward land.They increase precipitation on the windward sides of mountain ranges.They decrease precipitation totals on the leeward sides of mountain ranges.

Answer:

The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.

Explanation:

Because the heat of the plate will be releases warming up the air making it move faster

Answer:

0.036 moles of gas are contained in 890.0 mL at 21.0 C and 0.987 atm

Explanation:

Ideal gases are those gases whose molecules do not interact with each other and move randomly.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

where P represents the pressure of the gas, V its volume, n the number of moles of gas (which must remain constant), R the constant of the gases and T the temperature of the gas.

In this case:

Replacing:

0.987 atm* 0.890 L= n* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 294 K

Solving:

[tex]n=\frac{0.987 atm*0.890 L}{0.082\frac{atm*L}{mol*K}*294K }[/tex]

n= 0.036 moles

0.036 moles of gas are contained in 890.0 mL at 21.0 C and 0.987 atm

Answer:

Cathode

Cu^2+(aq) + 2e ----> Cu(s)

Zinc is the cathode

Explanation:

The plating of copper is normally done by electrolysis. Electrolysis is generally defined as the chemical decomposition produced by passing an electric current through a liquid or solution containing ions.

There are two electrodes, the anode and the cathode. Recall that electrolysis is not a spontaneous process, hence energy from a battery is required to drive the reaction in the desired direction.

The metal to be plated is normally the cathode while the metal used to plate it is normally the anode. Since copper is to be plated on zinc, zinc must be the cathode while copper will be the anode.

The half-reaction that takes place when pennies are plated with solid copper is :

Cu^2+(aq) + 2e ----> Cu(s)

Zinc is the cathode.

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Answer:

[tex]\Delta U = 244\,J[/tex]

Explanation:

The change in internal energy is given by the following expression:

[tex]\Delta U = n \cdot \bar c \cdot \Delta T[/tex]

[tex]\Delta U = (1\,mole)\cdot \left(24.4\,\frac{J}{mole\cdot K} \right)\cdot (10\,K)[/tex]

[tex]\Delta U = 244\,J[/tex]

Answer:

40g/mol

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Volume (V) = 500mL = 500/1000 = 0.5L

Mass of gass = 1g

Temperature (T) = – 23°C = – 23°C + 273 = 250K

Pressure (P) = 105 KPa = 105/101.325 = 1.04 atm

Number of mole (n) =?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the number of mole of the gas.

With the ideal gas equation, the number of mole of the gas can be obtained as follow:

PV = nRT

Divide both side by RT

n = PV / RT

n = (1.04 x 0.5)/(0.082 x 250)

n = 0.025mole

Step 3:

Determination of the molar mass of the gas:

This is illustrated below:

Mass of the gas = 1g

Number of mole of the gas = 0.025mole

Molar Mass of the gas =..?

Number of mole = Mass /Molar Mass

Molar Mass = Mass /number of mole

Molar Mass of the gas = 1/0.025

Molar Mass of the gas = 40g/mol

Therefore, the molar mass of the gas is 40g/mol

Answer: tap water to ice cube

Explanation:

Answer: tap water to ice cube

Explanation:

Answer:

Approximately [tex]1.3\; \rm mol \cdot L^{-1}[/tex]. (Assuming that the question says [tex]0.867[/tex] moles of salt in this [tex]0.69\; \rm L[/tex] solution.)

Explanation:

The molarity of a solution gives the quantity of the solute in every unit volume of the solution. In this question:

Note that in this question, liter is the unit for the volume of the solution. The molarity of the solution should thus give the amount of solute in every liter of the solution:

[tex]\begin{aligned} c & = \frac{n(\text{solute})}{V(\text{solution})} \\ &= \frac{0.867\; \rm mol}{0.69\; \rm mol} \approx 1.3\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Answer:

pH at the equivalence point for titration of HF and HCl will be basic and neutral respectively.

Explanation:

pH at equivalence point depends on hydrolysis equilibrium of conjugated base present in mixture.

[tex]\Rightarrow[/tex] Neutralization reaction: [tex]HF+OH^{-}\rightleftharpoons F^{-}+H_{2}O[/tex]

    Hence, at equilibrium, [tex]F^{-}[/tex] is present in mixture.

    Hydrolysis reaction: [tex]F^{-}+H_{2}O\rightleftharpoons HF+OH^{-}[/tex]

Here HF is an weak acid and [tex]OH^{-}[/tex] is a strong base, So, resultant pH of the           solution will be basic.

[tex]\Rightarrow[/tex] Neutralization reaction: [tex]HCl+OH^{-}\rightleftharpoons Cl^{-}+H_{2}O[/tex]

    Hence, at equilibrium, [tex]Cl^{-}[/tex] is present in mixture.

    Hydrolysis reaction: [tex]Cl^{-}+H_{2}O\rightleftharpoons HCl+OH^{-}[/tex]

Here HCl is a strong acid and [tex]OH^{-}[/tex] is a strong base, So, resultant pH of the           solution will be neutral.

Final answer:

At the equivalence point, titration of HCl with NaOH results in a neutral pH, while titration of HF with NaOH results in a basic pH.

Explanation:

To predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral, we need to consider the nature of the acid and base involved in the reaction. For the titration of a strong acid like hydrochloric acid (HCl) with a strong base like sodium hydroxide (NaOH), the equivalence point occurs at a pH of 7.00, resulting in a neutral solution. This is due to the formation of water from the neutralization reaction between HCl and NaOH.

However, for the titration of a weak acid like hydrofluoric acid (HF) with a strong base like NaOH, the equivalence point will be at a pH greater than 7, resulting in a basic solution. This occurs because the salt formed from the reaction of HF with NaOH gives a solution of sodium fluoride, NaF, which is basic due to the hydrolysis of F- anions in water.

Therefore, the pH of the equivalent point will be neutral for the titration of HCl with NaOH, and basic for the titration of HF with NaOH. The correct prediction for the titrations in question is that the equivalence point will be neutral for HCl and basic for HF.

Answer:

40659.855 J

Explanation:

From the question given above, we obtained the following:

Mass (m) = 18.015g

Heat of vaporisation (ΔHv) = 2257 J/g

Heat (Q) =?

The heat required to vaporise the water can be calculated as follow:

Q = mΔHv

Q = 18.015 x 2257

Q = 40659.855 J

Therefore, the heat required to vaporise the water is 40659.855 J

Answer

A. 8

Explanation:

The Atomic number is equal to the number of protons.

I took the test and got it right so this is 100% correct. It is NOT 18 like some people say

The atomic number of an oxygen atom with 8 protons is 8, regardless of the number of neutrons.

The atomic number of an atom is determined by the number of protons in its nucleus. This number also sorts elements into their correct position on the Periodic Table. Therefore, an oxygen atom with 8 protons will have an atomic number of 8, irrespective of the number of neutrons it has.

This is because neutrons do not influence the atomic number, only the atomic mass. So the correct answer to your question is: A. 8.

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Answer:

0.497 V

Explanation:

We need to apply the Nernst equation here. According to the Nernst equation;

Ecell= E°cell - 0.0592/n log Q

Where;

Ecell= emf of the cell under the given conditions

E°cell= standard emf of the cell

n= number of electrons transferred

Q= reaction quotient= [products]/[Reactants]= [Cr^2+]/[Fe^2+]

Balanced redox reaction equation; Cr(s)+Fe2+(aq)---------->Cr2+(aq)+Fe(s)

Values of standard electrode potential

Fe II: -0.44 V

Cr II: -0.91 V

E°cell= (-0.44) - (-0.99)

E°cell= 0.55V

[Fe2+ ]=0.0140M

[Cr2+ ]=0.862 M

Number of electrons transferred (n)= 2

Substituting into the Nernst's equation;

Ecell= 0.55- 0.0592/2 log [0.862]/[0.0140]

Ecell= 0.55 - 0.053

Ecell= 0.497 V

Answer:

The volume of the basketball at 273 K is 6.461 L

Explanation:

When the gas temperature increases, the molecules move faster and take less time to reach the walls of the container. This means that the number of crashes per unit of time will be greater. That is, there will be an increase in the pressure inside the container and the volume will increase.

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. So, the ratio between volume and temperature will always have the same value:

[tex]\frac{V}{T} =k[/tex]

In this case, you know:

Replacing:

[tex]\frac{7.1 L}{300K} =\frac{V2}{273 K}[/tex]

Solving:

[tex]V2=\frac{7.1 L}{300K} *273 K[/tex]

V2= 6.461 L

The volume of the basketball at 273 K is 6.461 L

According to Charles's law, the volume of a gas is directly proportional to its temperature at constant pressure. Using the formula V1/T1 = V2/T2, we can calculate the volume of the basketball at 273 K.

According to Charles's law, the volume of a gas is directly proportional to its temperature at constant pressure. We can use the formula V1/T1 = V2/T2 to solve this problem.

Given:

Using the formula, we can substitute the values and solve for V2:

V1/T1 = V2/T2

7.1 L / 300 K = V2 / 273 K

Cross multiplying, we get:

V2 = (7.1 L * 273 K) / 300 K

V2 = 6.46 L (rounded to two decimal places)