Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.3 g of methane is mixed with 52.3 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

Answer:

Explanation:

Hi there,

To get started, recall the Heat-specific heat capacity equation of a substance:

Q=mCΔT

Our final temperature is 47.8 °C, since this is the way it is worded in the response (from temperature X to temperature Y)

Quite simply, we can go ahead and plug in mass, specific heat capacity, and change in temperature as all the units match up!

[tex]Q= (525 g)(0.21 \ cal\ / g*C^{o} )(47.8C^{o}-13.0 C^{o})=3836.7 \ cal[/tex]

Study well and persevere.

thanks,

Final answer:

3823.8 calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C, calculated using the formula Q = mcΔT with a specific heat of 0.21 cal/g°C.

Explanation:

To find how many calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C, we use the formula for heat transfer Q = mcΔT, where 'm' is the mass of the substance, 'c' is the specific heat of the substance, and ΔT is the change in temperature.

First, we calculate the change in temperature (ΔT): ΔT = final temperature - initial temperature = 47.8°C - 13.0°C = 34.8°C. Using the provided values: m (mass) = 525g c (specific heat of Aluminum) = 0.21 cal/g°C ΔT (change in temperature) = 34.8°C

We plug these values into the equation to find the heat (Q): Q = m·c·ΔT Q = (525g)(0.21 cal/g°C)(34.8°C) Q = 3823.8 calories.

Therefore, 3823.8 calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C.

Answer:

The metal atoms that plate your object come from out of the electrolyte, so if you want to copper plate something you need an electrolyte made from a solution of a copper salt, while for gold plating you need a gold-based electrolyte—and so on

Explanation:

Plz give brainiest

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Answer:

electrolyte which is made from a solution of copper salt

Explanation:

Answer:

1.38×10^25 molecules

Explanation:

Applying n= (no. of molcules)/NA

23 = N/6.02×10^23

= 1.38×10^25 molecules

Answer:

Explanation:

Given parameters:

number of moles of CH₄  = 6.816 moles

number of moles of C₂H₆  = 0.589 mole

number of moles of C₃H₈  =  0.381 moles

Total pressure of the gases  = 3.93 atm

Unknown:

Partial pressure of the gases = ?

Solution:

Dalton's law of partial pressure states that " the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".

                         P[tex]_{T}[/tex]   = P[tex]_{1}[/tex]  + P[tex]_{2}[/tex]  + P[tex]_{3}[/tex]...............

where P[tex]_{T}[/tex]  = total pressure of the gas mixture

           P₁, P₂, P₃......  = partial pressure of gas 1, 2, 3............

The partial pressure of a gas is the pressure it would exert if confined alone in the volume of the gas mixture.

      Partial pressure of gas  = mole fraction of gas x total pressure of mixture

Now let us solve the problem.

Total number of moles of gases  = 6.816 moles + 0.589 mole + 0.381 moles

                                                        = 7.786moles

Partial pressure of CH₄ = [tex]\frac{6.816}{7.786} x 3.93 = 3.44 atm[/tex]

Partial pressure of C₂H₆  = [tex]\frac{0.589}{7.786}[/tex]  x 3.93  = 0.30atm

Partial pressure of C₃H₈  = [tex]\frac{0.381}{7.786}[/tex] x 3.93  = 0.19atm

Answer: 80mL

Explanation:

Its on ck12

5*16=80

To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to add 75 mL of water to the 5.00 mL of the stock solution.

To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to dilute the stock. She should add water to the 5.00 mL of the stock solution. Let's calculate the volume of water she needs to add.

First, we can use the equation (stock concentration) × (stock volume) = (final concentration) × (final volume) to find the volume of the diluted solution. Plugging in the values, we have (16.0 M) × (5.00 mL) = (1.00 M) × (final volume). Solving for the final volume, we get:

final volume = (16.0 M × 5.00 mL) / 1.00 M

final volume = 80 mL

To find the volume of water Maria needs to add, we subtract the volume of the stock solution from the final volume:

volume of water = final volume - volume of stock solution = 80 mL - 5.00 mL = 75 mL

Therefore, Maria needs to add 75 mL of water to the 5.00 mL of 16.0 M HCl solution to prepare a 1.00 M HCl solution.

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Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

[tex]C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}[/tex]

The ICE table is then shown as:

                               [tex]C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}[/tex]

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

[tex]K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}[/tex]

where ;

[tex]K_a = 3.02*10^{-5}[/tex]

[tex]3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}[/tex]

Since the value for [tex]K_a[/tex] is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

[tex]3.02*10^{-5} *(1.8) = {(x)(x)}[/tex]

[tex]5.436*10^{-5}= {(x^2)[/tex]

[tex]x = \sqrt{5.436*10^{-5}}[/tex]

[tex]x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M[/tex]

Dissociated form of  4-chlorobutanoic acid = [tex]C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M[/tex]

Percentage dissociated = [tex]\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100[/tex]

Percentage dissociated = [tex]\frac{7.3729*10^{-3}}{1.8 }*100[/tex]

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

Answer:

0.00091%

Explanation:

The fraction of 4-chlorobutanoic acid that would dissociatr in aqueos solution is a function of Ionization percentage. It is obtained by

Ka = [dissociated acid] / [original acid] x 100%

The equation of the reaction is

C₂HClCOOH  + H20 (aq)  = C₂HClCOO⁻  + H₃O⁺     pka =4.69

But pKa = - log Ka

   4.69  = - Log Ka

10⁻⁴⁶⁹   =    Ka

Taking the antilog of the equation we get the ionization constant

                    Ka   = 0.0000204 M

                           =2.04 x 10⁻⁵M

At the beginning of the reaction we have the following concentrations

1.8M C₂HClCOOH  :   0M (zero molar) C₂HClCOO⁻  :  0M (zero molar) H₃O⁺

At equilibrium, we have,

(1.8M -x)                                         xM C₂HClCOO⁻   and    xM H₃0⁺

Therefore,

Ka  =  [C2HClCOO-] [H30+] / [C2HClCOOH],

Inputing the value of Ka

                  Ka .[C2HClCOOH] = [CHCLCOO-] [H3O+]

                0.0000204 (1.8 - X) =  (x).(x)

                                            x² =  (0.00003672 - 0.0000204 X)

                                                 =  (3.672 x10-5 - 2.04 x10⁻⁵X)

                    x² +2.04x10⁻⁵ x     =  3.672 x 10⁻⁵

x² + 2.04 x 10⁻⁵x - 3.672 x 10⁻⁵ = 0

                                           x      = 0.00001632

                                                   = 1.632 x 10⁻⁵

Inputing back into equation 1

1.8 - x                  = [H3O+]

1.8 - 0.00001632 = 1.7999837

It therefore implies that only 0.00001632M of 4-chloroutanoic acid dissciated at equilibrium, we can now calculate the percentage dissociation by

Percentage dissociation = 0.00001632 / 1.8M x 100%

                                         = (1.632 x 10⁻⁵/1.8 ) x 100%

                                         = 0.00090667%

                                          = 0.00091%

Answer:

11.88

Explanation:

The capacity of container to be able to hold any amount of moles is dependent on size of container.

Mole is defined as the unit of amount of substance . It is the quantity measure of amount of substance of how many elementary particles are present in a given substance.

It is defined as exactly 6.022×10²³ elementary entities. The elementary entity can be a molecule, atom ion depending on the type of substance. Amount of elementary entities in a mole is called as Avogadro's number.

It is widely used in chemistry as a suitable way for expressing amounts of reactants and products.For the practical purposes, mass of one mole of compound in grams is approximately equal to mass of one molecule of compound measured in Daltons. Molar mass has units of gram per mole . In case of molecules, where molar mass  in grams present in one mole  of atoms is its atomic mass.

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Answer:

278 balloons can be aired.

Explanation:

We apply the Ideal Gases Law to determine the total available volume for each baloon of 0.75L. So we need to divide the volume we obtain by the volume of each baloon.

We determine the moles of CO₂

375 g. 1 mol / 44g = 8.52 moles

V = (n . R . T) / P → (8.52 mol . 0.082 . 298K) / 1 atm = 208.3 L

That's the total volume from the tank, so we can inflate (208.3 / 0.75) = at least 278 balloons

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

First, we will convert 375 g of CO₂ to moles using its molar mass (44.01 g/mol).

[tex]375 g \times \frac{1mol}{44.01g} = 8.52 mol[/tex]

1 mole of CO₂ at 1.0 atm and 298 K occupies 24.4 L. The volume occupied by 8.52 moles of CO₂ is:

[tex]8.52 mol \times \frac{24.4L}{mol} = 208 L[/tex]

8.52 moles of CO₂ at 1.0 atm and 298 K occupies 208 L. The number of 0.75 L-balloons that can be filled is:

[tex]208 L \times \frac{1balloon}{0.75L} \approx 277 balloon[/tex]

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

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The reaction will consume 3 moles of iron(III) oxide and will produce 3 moles of aluminum oxide and 12 moles of iron, following the mole ratio from the balanced chemical equation.

Starting from the balanced chemical equation, we can see that the ratio of "Fe2O3" (iron(III) oxide) to "Al" (aluminum) is 1:2. This ratio allows us to calculate the amount of moles that will react or produce. We can apply these ratios in the following way:

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Answer:

Nuclear reaction takes place at the nucleus whereas chemical reaction involves valence electrons

Explanation:

Answer:

4.77 is the pH of the given buffer .

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})[/tex]

We are given:

[tex]K_a[/tex] = Dissociation constant of propanoic acid = [tex]1.3\times 10^{-5}[/tex]

[tex][CH_3CH_2COONa]=0.254 M[/tex]

[tex][CH_3CH_2COOH]=0.329 M[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})[/tex]

pH = 4.77

4.77 is the pH of the given buffer .

Final answer:

To determine the pH of a buffer with given concentrations of propanoate and propanoic acid, apply the Henderson-Hasselbalch equation using the given Ka value for propanoic acid and the concentrations of the acid and its conjugate base.

Explanation:

To calculate the pH of a buffer consisting of 0.254 M sodium propanoate (CH3CH2COO-Na+) and 0.329 M propanoic acid (CH3CH2COOH), we use the Henderson-Hasselbalch equation:

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

Where:

First, we calculate the pKa:

pKa = -log(Ka) = -log(1.3 x 10-5)

Then plug the values into the Henderson-Hasselbalch equation:

pH = pKa + log(0.254/0.329)

After calculating the above expression, we find the pH of the buffer solution.

This is an incomplete question, here is a complete question.

The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?

[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]

Answer : The value of [tex]K_p[/tex] is, [tex]5.79E^{-7}[/tex]

Explanation :

The relation between [tex]K_c[/tex] and [tex]K_p[/tex] is:

[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]

where,

[tex]K_p[/tex] = equilibrium constant at constant pressure  = [tex]3.3\times 10^{-5}[/tex]

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

T = temperature = 700 K

[tex]\Delta n[/tex] = change in number of gaseous moles = Product moles - Reactant moles = (2+1) - (2) = 3 - 1 = 1 mol

[tex]K_c[/tex] = equilibrium constant

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

[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]

[tex]3.3\times 10^{-5}=K_c(0.0821\times 700)^{1}[/tex]

[tex]K_c=5.7\times 10^{-7}=5.7E^{-7}[/tex]

Therefore, the value of [tex]K_p[/tex] is, [tex]5.7E^{-7}[/tex]

The value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]

The given parameters;

The change in the number of gaseous moles is calculated as follows;

[tex]\Delta n = product \ - reactant\\\\\Delta n = (2+1) - 2 = 1 \ mole[/tex]

The value of the equilibrium constant (Kc) is calculated as follows;

[tex]K_p = K_c \times (RT)^{\Delta n}\\\\3.3 \times 10^{-5} = K_c \times (0.0821 \times 700)^1\\\\3.3 \times 10^{-5} = 57.47K_c\\\\K_c = \frac{3.3 \times 10^{-5} }{57.47} \\\\K_c = 5.74 \times 10^{-7} \[/tex]

Thus, the value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]

"Your question is not complete, it seems to be missing the following information;"

The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?

[tex]2SO_3(g) \ ---> \ 2SO_2(g) \ + \ O_2[/tex]

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Answer:

London dispersion forces

Explanation:

The London dispersion force is the weakest kind of intermolecular force. The London dispersion force is a temporary attractive force that occurs when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction.

These London dispersion forces are mostly seen in the halogens (e.g., Cl2 and I2), the noble gases (e.g., Xe and Ar), and in many non-polar molecules, such as carbon dioxide and propane. London dispersion forces are part of the van der Waals forces, and are very weak intermolecular attractions.

Answer:

When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.

Explanation:

Ba²⁺(aq) + 2 F⁻(aq) <----> BaF₂(s)

When BaF₂ precipitates, the Ksp relation is given by

Ksp = [Ba²⁺] [F⁻]²

[Ba²⁺] = 0.0144 M

[F⁻] = ?

Ksp = (1.7 × 10⁻⁶)

1.7 × 10⁻⁶ = (0.0144) [F⁻]²

[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555

[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M

Hope this Helps!!!

When the concentration of F⁻ exceeds 0.0109 M in a solution containing 0.0144 M Ba²⁺, BaF₂ will begin to precipitate.

To determine when BaF₂ will precipitate from a solution containing Ba²⁺ and F⁻ ions, we need to use the solubility product constant (Ksp). For BaF₂, Ksp is given as 1.7 × 10⁻⁶.

The dissociation reaction for BaF₂ is:

BaF₂ (s) ⇌ Ba²⁺ (aq) + 2F⁻ (aq)

The Ksp expression is:

Ksp = [Ba²⁺][F⁻]²

Given [Ba²⁺] = 0.0144 M, we need to find [F⁻] when the solution just starts to precipitate.

Therefore, when the concentration of F⁻ exceeds approximately 0.0109 M, BaF₂ will start to precipitate.

Answer:

0.1 mol O2

Explanation:

1 mol of any gas at STP = 22.4 L

2.24L/22.4L = 0.1

1mol * 0.1 = 0.1 mol

you have 0.1 mol of O2

The number of moles of [tex]O_2[/tex]  is required.

The number of moles of [tex]O_2[/tex] is 0.2 moles.

V = Volume = 2.24 L

[tex]\rho[/tex] = Density of [tex]O_2[/tex] at STP = 1.429 g/L

M = Molar mass of [tex]O_2[/tex] = 16 g/mol

Density is given by

[tex]\rho=\dfrac{m}{V}\\\Rightarrow m=\rho V\\\Rightarrow m=1.429\times 2.24\\\Rightarrow m=3.20096\ \text{g}[/tex]

Molar mass is given by

[tex]M=\dfrac{m}{n}\\\Rightarrow n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{3.20096}{16}\\\Rightarrow n=0.2\ \text{moles}[/tex]

The number of moles of [tex]O_2[/tex] is 0.2 moles.

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The student can check the purity of the CaBr2 sample by performing a quantitative chemical analysis to compare the theoretical and actual mass of the compound, which will reveal the presence of any impurities.

A student can determine whether the 10.0g sample labeled CaBr2 is pure by calculating the molar mass of CaBr2 and comparing the theoretical mass to the actual mass of the sample. If there is a discrepancy between the theoretical and actual mass, it could indicate the presence of an inert impurity.

To carry out the analysis, the student must perform a quantitative chemical analysis, such as a titration, to determine the amount of active ingredient in the sample. The student could also measure the mass of CaBr2 after a precipitation reaction, as described in the problem statement. If the resulting mass corresponds to the known molar mass of a pure CaBr2 sample, then the sample can be considered pure. On the other hand, if the mass deviates from the expected value, then an impurity might be present.

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

The chemical equation for this reaction is:

HA  +  B  ⇌  A⁻  +  HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Reason: In this reaction, there is no exchange of proton between the acid and the base.

Answer: The molecular weight of this compound is 288.4 g/mol

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]

where,

[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

[tex]x_2[/tex] = mole fraction of solute  =[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]

Given : 7.745 g of compound is present in 159.9 g of diethyl ether

moles of solute = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.745g}{Mg/mol}[/tex]

moles of solvent (diethyl ether) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{159.9g}{74.12g/mol}=2.157moles[/tex]

Total moles = moles of solute + moles of solvent  = [tex]\frac{7.745g}{Mg/mol}+2.157[/tex]

[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}[/tex]

[tex]\frac{463.57-457.87}{463.57}=1\times \frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}[/tex]

[tex]M=288.4g/mol[/tex]

Thus the molecular weight of this compound is 288.4 g/mol

The molar mass of the unknown gas which has a mass of 36.5 g is 146 g/mol

Mass of He = 0.5 g

Molar mass of He = 4 g/mol

At standard temperature and pressure (stp),

1 mole of He = 22.4 L

Therefore,

0.125 mole of He = 0.125 × 22.4

0.125 mole of He = 2.8 L

Volume of He = 2.8 L

Volume of unknown gas =?

Volume of He = ½ × Volume of unknown gas

2.8 = ½ × Volume of unknown gas

Cross multiply

Volume of unknown gas = 2.8 × 2

Volume of unknown gas = 5.6 L

22.4 L = 1 mole of unknown gas

Therefore,

5.6 L = [tex]\frac{5.6}{22.4}\\\\[/tex]

5.6 L = 0.25 mole of unknown gas.

Mole of unknown gas = 0.25 mole

Mass of unknown gas = 36.5 g

Molar mass of unknown gas =?

[tex]Molar mass = \frac{mass}{mole} \\\\ = \frac{36.5}{0.25}\\\\[/tex]

Therefore, the molar mass of the unknown gas is 146 g/mol

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At STP, one mole of any ideal gas occupies around 22.4 L. As the 0.500 g Helium sample occupies half the volume of the unknown gas, it means the that the molar mass of the unknown gas would be twice the mass of the helium. The molar mass of the unknown gas is calculated to be 292 g/mol.

In order to determine the molar mass of the unknown gas, we need to observe the characteristics of gases at Standard Temperature and Pressure (STP). At STP, one mole of any ideal gas will occupy approximately 22.4 L. Given that the 0.500 g sample of Helium with a molar mass of 4 g/mol occupies half of the volume the unknown gas does, we can conclude that one mole of the unknown gas will have twice the mass of the Helium sample.

So, if 0.500 g of He occupies a volume that is equal to the volume of 0.125 moles of He (since 4 g of He = 1 mole), this means the same volume is being occupied by the unknown gas. Given the mass of the unknown gas is 36.5 g, the molar mass of the unknown gas would be 36.5 g / 0.125 moles = 292 g/mol.

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Answer:

Because the hot air from the equator is balance with the cold air from the polar region, meaning the temperature is the right degree, therefore it causes the slowing down of that hurricane.

Explanation:

From your science class you do study the convectional current right? that's what happen on the outside real life

As more than just a result of the equator's hot air being in balance with the poles' cold air, which means the temperature is at the right level, the hurricane is slowed down.

Hurricanes are low-pressure systems with organized thunderstorm activity that originate over tropical or subtropical waters. They are sometimes referred to as tropical cyclones in general. The warm ocean waves provide them energy.

When warm, humid air begins to rise over water, hurricanes emerge. Cooler air replaces the rising air. Large clouds and thunderstorms continue to grow as a result of this process. Thanks to the earth's Coriolis Effect, these thunderstorms are still expanding and starting to rotate.

The greatest wind speeds during Katrina's impact near Grand Isle, Louisiana, may have reached 140 mph. The NWS Doppler Radar at Mobile (KMOB) recorded winds of up to 132 mph between 3,000 and 4,000 feet above ground level in the morning as Katrina moved farther north and made a second impact along the Mississippi/Louisiana border.

Thus, the equator's hot air being in balance with the poles' cold air, which means the temperature is at the right level, the hurricane is slowed down.

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The equilibrium constant Kc for the reaction Co(s) + [tex]CO_{2}[/tex](g) ---->  CoO(s) + CO(g) at 823°C is the reciprocal of Kp for the reverse reaction, resulting in a Kc of approximately 2.04 x 10^-3.

The value of the equilibrium constant Kc for the reaction Co(s) + [tex]CO_{2}[/tex](g) --------> CoO(s) + CO(g) can be calculated using the relationship between Kp and Kc and the fact that the reaction is the reverse of the one given. For the given reaction, Kp = 490 at 823°C. Since there are the same number of moles of gas on each side of the reaction, the relationship between Kp and Kc does not require any correction for the change in moles of gas, so Kc is the reciprocal of Kp for the reverse reaction. Thus, Kc for Co(s) + [tex]CO_{2}[/tex] ------->CoO(s) + CO(g) is 1/490 or approximately 2.04 x 10^-3 at 823°C.

Final answer:

To prepare the buffer, you would need to add 2.808 grams of Na2EDTA · 2H2O to the 500 mL volumetric flask containing 1.97 grams of Ba(NO3)2. This will give the desired buffer with a pBa2+ of 7.00 at pH 10.00.

Explanation:

To calculate the mass of Na2EDTA · 2H2O needed to prepare the buffer, we need to use the equation:

Kf' = αY4− · Kf = [MYn−4][Mn+][EDTA]

Given that log Kf = 7.88 and αY4− = 0.30, we can rearrange the equation to solve for the concentration of MYn−4:

[MYn−4] = Kf' / (αY4− · Kf) = 10^(7.00 − 7.88) / (0.30 · 10^7.88) = 0.562 M

Now, we can calculate the moles of Ba(NO3)2:

moles of Ba(NO3)2 = mass / molar mass = 1.97 g / 261.35 g/mol = 0.00754 mol

Since the stoichiometric ratio between Na2EDTA · 2H2O and Ba(NO3)2 is 1:1, the moles of Na2EDTA · 2H2O required is also 0.00754 mol.

Finally, we can calculate the mass of Na2EDTA · 2H2O:

mass = moles × molar mass = 0.00754 mol × 372.23 g/mol = 2.808 g

It shows the neutralization property of the base and there is no pink color.

Explanation:

CuSO₄(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)

Here NaOH is the base and Copper sulfate is an salt of an acid and so it is acidic salt. When these two compounds react, we will get the precipitate of copper hydroxide and Sodium sulfate.

And this reaction seems to be a neutralization reaction, since an acidic salt is neutralized by a base.

Since NaOH is converted into salt, that is sodium sulfate (Na₂SO₄), the phenolphthalein indicator becomes colorless in this reaction. It doesn't turn pink.

Answer:

Check the explanation

Explanation:

A. Calcite ------> Argonite

At 298 K and 1 atm,

\DeltaGrxn = \DeltaGarg - \DeltaGcalc

= -1127.71-(-1128.76) = 1.05 kJ/mol

\DeltaGrxn is positive, so the reaction is not spontaneous.

Thus, the reaction would not happen even if enough time is given.

B. Molar mass of CaCO3 = 100 g/mol

Volume = mass/density

Vcalc = 100/2.710 = 36.900 ml/mol = 36.9*10-3 L = 36.9*10-6 m3

Varg = 100/2.930 = 34.130 ml/mol = 34.13*10-3 L = 34.13*10-6 m3

\DeltaGrxn = 1.05 kJ/mol = 1.05*103 J/mol

\DeltaG = -\DeltaP\DeltaV

\DeltaP = -\DeltaG/\DeltaV = -1.05*103/(34.13*10-6 - 36.9*10-6) = 0.379*109 Pa = 3.79*108 Pa

The pressure should be more than 3.79*108 Pa.

C. For reaction, \DeltaSrxn = Sarg - Scalc

= 88.70-92.88 = -4.18 J/mol.K

\DeltaHrxn = -1207.04-(-1206.87) = -0.17kJ/mol

At 400 K, \DeltaG = \DeltaH-T\DeltaS

= -0.17-(400*-4.18)

= +ve

So, the reaction is not favorable on increasing the temperature to 400 K.

D. \DeltaT = 400-298 = 102 K

\DeltaP = 1000-1 = 999.0 atm = 999*106 Pa

\DeltaG = \DeltaS\DeltaT - \DeltaV\DeltaP

= (-4.18*102)-(-2.77*10-6*999*106)

= -426.36+2767.23 = 2340.87 J/mol

\DeltaG = +ve

calcite would be more stable.

Answer: PLease make me Brainliest

"They can bring anything from tropical warm and humid days to arctic cold depending on the type of air mass. Fronts form the boundaries of air masses with differing properties. The most severe weather usually occurs when dry-cold continental polar air clashes with warm-humid maritime tropical air."

Final answer:

When a continental polar air mass meets a continental tropical air mass, the interaction can create a weather front that often leads to precipitation and can spawn extratropical cyclones, resulting in variable weather conditions including rain, snow, and thunderstorms.

Explanation:

If a continental polar air mass (cP) clashed with a continental tropical air mass (cT), a significant weather event would likely occur. The cP air mass is characterized as cold and dry, and is a stable air mass that can bring cooler temperatures to affected regions. On the other hand, a cT air mass is hot and dry. When these two air masses meet, the differing temperatures and densities of the air can cause the formation of a weather front, often leading to precipitation and potentially stormy conditions.

The collision often happens along what's called the polar front, where a cyclonic shear is created due to the opposing streams of air. This interaction can lead to the formation of extratropical cyclones, which are large low-pressure systems that can cause a wide array of weather conditions, including rain, snow, thunderstorms, and in some cases, severe weather outbreaks.

Given that cP air is denser than cT air, the warmer, less dense cT air would rise above the cP air mass. This rising motion can lead to cooling and condensation of water vapor, resulting in cloud formation and precipitation. This is a key mechanism through which weather disturbances are generated in regions where such air masses clash, particularly in the midlatitudes, which are notorious for their variable weather patterns.

Final answer:

The addition of solid NaOH to a reaction can decrease the amount of work a chemical system can perform by shifting the equilibrium position and altering the concentration of reactants and products involved in work

Explanation:

The statement that will decrease the amount of work the system could perform is (b) Add solid NaOH to the reaction (assume no volume change). In a chemical system, work is related to changing conditions that affect the direction of equilibrium according to Le Chatelier's principle. Adding solid NaOH to a reaction mixture could shift the equilibrium position in a way that may result in the consumption of reactants or products involved in performing work (such as electrical or mechanical work in an electrochemical cell or in muscle contraction). For example, if the reaction were exergonic and NaOH was a product, adding more of it would shift the equilibrium to the left, thereby reducing the amount of work the system could do. This is because there would be a greater proportion of reactants and less of the energy-containing products required to perform work.

Answer:

1.surface area of a solid reactant.2.concentration or pressure of a reactant.

3temperature.

4nature of the reactants.

5presence/absence of a catalyst.

Explanation:

The five main factors that can affect rates of a chemical reaction are concentration of reactants, temperature, pressure, presence of a catalyst, and surface area of solid reactants.

The rate of a chemical reaction can be influenced by several factors, including:

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Answer:

One

Explanation:

Because it reaches an equibrum state so it's equals to one

Answer:

pH= 7

Explanation:

pOH=-log [OH-] = -log (10^-7)= 7

pH= 14-pOH= 14-7

=7

Answer:

vhgbvbhdf

Explanation:

(a) It is given that the gas is ideal. Formula for change in entropy of the gas according to the first law of thermodynamics is as follows.

       [tex]\Delta U = dQ - dW[/tex]

For isothermal process, [tex]\Delta U = 0[/tex] at constant temperature.

So,   [tex]\Delta U = dQ - dW[/tex]

      [tex]0 = dQ - dW[/tex]

or,       dQ = dW = 730 J

Now, according to the second law of thermodynamics the entropy change is as follows.

           [tex]\Delta S = \frac{dQ}{dT}[/tex]

                        = [tex]\frac{730 J}{293.15 K}[/tex]

                        = 2.490 J/K

Therefore, the entropy change of the gas is 2.490 J/K.

(b)  In the given process, at constant temperature the gas will be compressed slowly because then kinetic energy of the gas molecules will also be constant. The volume decreases so that the movement of molecules increases as a result, entropy of molecules will also increase.

This means that [tex]\Delta S = 2.490 J/K > 0[/tex]

An ideal gas is a hypothetical gas in which there are no intermolecular attractions between the molecules of a gas. The collision between the molecules is perfectly elastic.

The answers are as follows:

(a) The ideal gas is given, in which the change of entropy of the gas can be calculated by the first law of thermodynamics.

[tex]\Delta[/tex] U = dQ - dW

In an isothermal process, the \Delta U = 0 at constant temperature, such that:

[tex]\Delta[/tex] U = dq - dW

0 = dQ - dW

dQ = dW = 730 J (given)

Now, based on the second law of thermodynamics, the entropy change will be:

[tex]\Delta S = \dfrac {\text {dQ}}{\text {dW}}\\\\\Delta S = \dfrac {730 \text J}{293.15 \text K}[/tex]

[tex]\Delta[/tex]S = 2.490 J/K

Thus, the entropy change for the gas is 2.490 J/K.

(b) At constant temperature, the gas will be compressed slowly because the kinetic energy of the molecules will be constant. The decrease in the volume will be due to the increase in the movement of the molecules, which will cause an increase in the entropy. This means [tex]\Delta[/tex] S will be greater than  2.490 J/K.

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