If 1 10 liter sample of O2 gas at 300 kelvin and 0.5 atmosphere of pressure contains 5.0 x 10^22 molecules ....... how many molecules would a 10 liter sample of H2 have if it was at the same temperature and pressure as the oxygen remember that hydrogens are 1/16th the size of oxygen

The molecules CF4 and NF3 have five electron groups arranged in a trigonal bipyramidal fashion. CF4, with four bonded atoms and zero lone pairs, is tetrahedral. NF3, with three bonded atoms and one lone pair, is trigonal pyramidal.

In CF4 and NF3, the five electron groups on the central C and N atoms have a trigonal bipyramidal arrangement. The shapes of the molecules are determined by the number of pairs of electrons: since CF4 has four bonded atoms and zero lone pairs of electrons, the shape is tetrahedral. Since NF3 has three bonded atoms and one lone pair of electrons, the shape is trigonal pyramidal.

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CF4 and NF3 have a tetrahedral electron-pair geometry due to four electron groups on the central atoms. CF4 has a tetrahedral shape due to four bonded atoms and no lone pairs, while NF3 has a trigonal pyramidal shape due to three bonded atoms and one lone pair.

In CF4 and NF3, the four electron groups on the central C and N atoms have a tetrahedral arrangement. The shapes of the molecules are determined by the number of bonded atoms and lone pairs of electrons:

This distinction occurs because although both molecules have a tetrahedral electron-pair geometry, the presence of a lone pair of electrons on nitrogen in NF3 distorts the geometry to be trigonal pyramidal at the molecular level.

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Answer:

Photosynthesis

Explanation:

As the plants are converting the CO2 into O2... So the photosynthesis reduce the amount of CO2...

Answer:

The answer is photosynthesis

Explanation:

The key processes of the carbon cycle are when the carbon dioxide from the atmosphere is changed into plant material in the biosphere by photosynthesis.

Then organisms in the biosphere obtain energy from respiration and releases carbon dioxide that was orignally changed by photosynthesis.

Answer:

pH of HCl solution is 1.44

Explanation:

NaOH is a monoprotic base and HCl is a monobasic acid.

Neutralization reaction: [tex]NaOH+HCl\rightarrow NaCl+H_{2}O[/tex]

According to balanced reaction, 1 mol of NaOH neutralizes 1 mol of HCl.

Number of moles of NaOH in 13 mL of 0.21 M NaOH

= [tex]\frac{0.21}{1000}\times 13mol=0.00273mol[/tex]

let's assume concentration of HCl is C (M)

Then, number of moles of HCl in 75 mL of C (M) HCl solution

= [tex]\frac{75}{1000}\times Cmol=\frac{75C}{1000}mol[/tex]

So, we can write, [tex]\frac{75C}{1000}=0.00273[/tex]

or, [tex]C=0.0364[/tex]

1 mol of HCl contains 1 mol of [tex]H^{+}[/tex]

So, concentration of [tex]H^{+}[/tex] in 0.0364 M HCl, [tex][H^{+}]=0.0364M[/tex]

Hence, [tex]pH=-log[H^{+}]=-log(0.0364)=1.44[/tex]

Final answer:

To balance the equation RbOH + H3PO4 → Rb3PO4 + H2O, you use the coefficients 3, 1, 1, and 3, resulting in a sum of 8.

Explanation:

The question involves balancing a chemical equation to ensure that the same number of atoms of each element is present on both sides, following the law of conservation of mass. The balanced equation for RbOH + H3PO4 → Rb3PO4 + H2O is 3RbOH + H3PO4 → Rb3PO4 + 3H2O, giving us coefficients 3, 1, 1, 3, and the sum of these coefficients is 8.

It's important to never change the subscripts in a chemical formula while balancing an equation. Instead, coefficients are placed in front of the chemical formulae to indicate the relative amounts of reactants and products. The lowest whole number ratio that balances the equation should be used.

The molar mass of the gas is 77.20 gm/mole.

Explanation:

The data given is:

P = 3.29 atm,   V= 4.60 L   T= 375 K  mass of the gas = 37.96 grams

Using the ideal Gas Law will give the number of moles of the gas. The formula is

PV= nRT    (where R = Universal Gas Constant 0.08206 L.atm/ K mole

Also number of moles is not given so applying the formula

n= mass ÷ molar mass of one mole of the gas.

n = m ÷ x   ( x  molar mass) ( m mass given)

Now putting the values in Ideal Gas Law equation

PV = m ÷ x RT

3.29 × 4.60 = 37.96/x × 0.08206 × 375

15.134 = 1168.1241  ÷ x

15.134x = 1168.1241

x = 1168.1241 ÷ 15.13

x = 77.20 gm/mol

If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.

Answer:

77.2

Explanation:

Explanation:

Answer:

increase in solvent temperature will increase the solubility of solid particle but decreases the solubility of gas particle.

Explanation:

In solid particle when temperature increases its help to break apart the solid particle and increase in kinetic energy of solvent results in increase in solubility of solid particles in solvent.

but in gas solute, increase in temperature of solvent causes the increase in motion of gas molecules means increase in kinetic energy of molecules in the gas which results in breakage of inter molecular bonds and removal of the molecules from the heated solution.

Answer:

One significant difference between gases and liquids is that a gas expands to fill its container.

Explanation:

One of the biggest differences between liquids and gases is that liquids have a defined volume and gases do not. A gas has no defined shape or volume, it acquires the shape and volume of the container in which it is located. It is the tendency of gases to increase in volume due to the force of repulsion that works between their molecules.

One significant difference between gases and liquids should be that a gas expands to fill its container.

It is the biggest difference that lies between the liquids and gases is that the liquid should contain the volume while the gas does not. Also, the gas does not have a defined shape or volume.

It purchased the shape and the container volume where it should be located.

Also, there is the gas tendency for increasing the volume because of the repulsion of the force that should be worked between the molecules.

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Answer: 2.53 x 10^-2 moles As ( the answer is in scientific notation )

Answer:

The concentration of NaOH in the final solution 0.0584 mole/lit

Explanation:

So Molarity of the solution after mixing [tex]=\frac{0.365}{6.245929} = 0.0584 (M)[/tex]

∴ The concentration of NaOH in the final solution 0.0584 mole/lit

The concentration of NaOH in the final solution is thus 0.0531 M.

To determine the concentration of NaOH in the final solution after 8.4 grams of NaOH are dissolved into 620 mL of 0.250 M NaOH solution, which is then mixed into 1.65 gallons of water, follow these steps:

First, calculate the moles of NaOH initially dissolved. Knowing NaOH has a molar mass of 40.0 g/mol, 8.4 g of NaOH translates to 0.21 moles of NaOH.

Add the moles from the initial 620 mL of 0.250 M solution. That will be 0.250 moles/L * 0.62 L = 0.155 moles.

Combine the moles from both sources for a total of 0.365 moles of NaOH.

Convert 1.65 gallons of water to liters (1 gallon = 3.78541 L), making 1.65 gallons equal to 6.245 liters. Adding this to the initial 0.62 L yields a total final volume of 6.865 liters.

Finally, calculate the final concentration by dividing the total moles of NaOH by the final volume of solution in liters to find the molarity of the final solution.

The concentration of NaOH in the final solution is thus 0.0531 M.

Answer:

123.1

Explanation:

The average atomic mass of an element which exhibits isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.

Average atomic mass = (95% of 123 amu)+(5% 0f 125 amu)

= (0.95 × 123) + (0.05 × 125)

= 116.85 + 6.25

= 123.1   (to 1 decimal place)

Answer:

The level of hydrogen ion has increased by a factor of 10.

Explanation:

We know, [tex]pH=-log[H^{+}][/tex]

where [tex][H^{+}][/tex] represents concentration of [tex]H^{+}[/tex] in molarity

Here [tex](pH)_{final}=6.5[/tex]

So, [tex][H^{+}]_{final}=10^{-6.5}M[/tex]

Here [tex](pH)_{initial}=7.5[/tex]

So, [tex][H^{+}]_{initial}=10^{-7.5}M[/tex]

Hence, change in pH = [tex]\frac{[H^{+}]_{final}}{[H^{+}]_{initial}}[/tex]  = [tex]\frac{10^{-6.5}M}{10^{-7.5}M}=10[/tex]

So, the level of hydrogen ion has increased by a factor of 10.

Option (B) is correct

Explanation:

Below are attachments containing the solution

Answer:

a. chain length and degree of saturation.

Explanation:

Fatty acids can be classified according to the length of their chain, for example, in short (if it has less than 8 carbons), medium (between 8-12 carbons), long (between 12-18 carbons) and very long (if it has less than 18 carbons); they are also classified according to their degree of unsaturation, in saturated, monounsaturated and polyunsaturated; and according to the isomerism in cis and trans fatty acids.

Fatty acids may differ from one another in chain length and degree of saturation. Therefore, the correct option is option A.

Fatty acid is a crucial part of lipids, which are the parts of living things including plants, animals, and microbes that may be dissolved in fat. A fatty acid normally consists of a chain that is straight with a even amount of carbon atoms, hydrogen atoms running along the entire length of the chain, a carboxyl group (COOH) at one end, and hydrogen atoms running the duration of the chain. It forms an acid (carboxylic acidic solution) because of the carboxyl group. The acid is saturated if there are only single carbon-to-carbon bonds. Fatty acids may differ from one another in chain length and degree of saturation.

Therefore, the correct option is option A.

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Final answer:

The average rate of the reaction is -0.005 M/min.

Explanation:

The average rate of a reaction can be calculated by using the change in concentration of a reactant or product divided by the time interval. In this case, the concentration of NO changed from 0.100 M to 0.025 M in 15 minutes. You can calculate the average rate of the reaction by dividing the change in concentration by the time interval:

Average rate = (Final concentration - Initial concentration) / Time

Average rate = (0.025 M - 0.100 M) / 15 minutes

Average rate = -0.075 M / 15 minutes = -0.005 M/min

Answer:

Explanation:

Sodium is a group 1 element with atomic number 1. It has 11 electrons. It is soft reactive metal. It has 1 valence electron.

Fluorine is a group 7 element, a hologen with 7 valence electron. It is a most reactive non metal.

When sodium react with fluorine, ionic bond is formed in the resulting compound sodium fluoride.

One sodium and fluorine each totaling 2 atoms are enough to make the bond.

As the bond is formed, both atoms have octet structure. That is they each have 8 electrons on their outermost shells.

The positive charge on sodium indicates that sodium had lost 1 electron to fluorine atom.

The negative charge on fluorine ion indicates that fluorine atom had gained 1 electron from sodium atom to form negative ion.

The name of the compound is sodium fluoride with formula NaF.

Answer:

The reaction of 1st group element Na and 7th group element F, results in formation of Sodium fluoride (NaF).

Explanation:

1. (a) Sodium (Na) is element belonging to the first group of the periodic table.        

(b) The atomic number of Na is 11 and after distribution it has 1 valence electron.

2. (a) Fluorine (F) is the halogen element and belongs to group 7 of the periodic table.

(b) The atomic number of F is 19 and it has 7 valence electrons.

3. The reaction of Na with F results in an ionic interaction with the formation of NaF (Sodium fluoride).

(a) The reaction and combination of Na and F is likely to be ionic in nature.

(b) The one valence electron of Na and F are involved in formation of bond. Total 2 electrons are involved.

4. (a) The successful formation of bond result in octet completion of NaF by sharing of electrons. Both the elements share 8 electrons, i.e. 7 electrons of F and 1 electron of Na.

(b) The +ve charge is imparted for the loss of electron . Na lost its 1 valence electron to F for bond formation and thus imparts the +ve charge.

(c) The -ve charge is imparted with the gain of electron. In reaction of Na with F, fluorine being more electronegative gains its electrons from Na and imparts a -ve charge.

(d) The reaction of Na with F results in the formation of compound known as sodium fluoride with the molecular formula NaF.

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Answer:

Work done in this process = 4053 J

Explanation:

Mass of the gas = 0.092 kg

Pressure is constant = 1 atm = 101325 pa

Initial temperature [tex]T_{1}[/tex] = 200 K

Final temperature [tex]T_{2}[/tex] = 200 - 85 = 115 K

Gas constant for nitrogen = 297 [tex]\frac{J}{kg k}[/tex]

When pressure of a gas is constant, volume of the gas is directly proportional to its temperature.

⇒ V ∝ T

⇒ [tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{T_{2} }{T_{1} }[/tex] ------------ ( 1 )

From ideal gas equation [tex]P_{1}[/tex] [tex]V_{1}[/tex] = m R [tex]T_{1}[/tex] ------ (2)

⇒ 101325 × [tex]V_{1}[/tex] = 0.092 × 297 × 200

⇒ [tex]V_{1}[/tex] = 0.054 [tex]m^{3}[/tex]

This is the volume at initial condition.

From equation 1

⇒ [tex]\frac{V_{2} }{0.054}[/tex] = [tex]\frac{200}{115}[/tex]

⇒ [tex]V_{2}[/tex] = 0.094 [tex]m^{3}[/tex]

This is the volume at final condition.

Thus the work done is given by W = P [[tex]V_{2}[/tex] - [tex]V_{1}[/tex] ]

⇒ W = 101325 × [ 0.094 - 0.054]

⇒ W = 4053 J

This is the work done in that process.

Answer:

[tex]\large \boxed{\text{7.7 kJ}\cdot\text{mol}^{-1} }[/tex]

Explanation:

1. Mass of solution

[tex]\text{Mass of solution} = \text{150 mL} \times \dfrac{\text{1.02 g}}{\text{1 mL}} = \text{153 g}[/tex]

2. Calorimetry

There are two energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

[tex]\begin{array}{ccccl}n\Delta H & + & mC\Delta T &=&0\\\text{0.70 mol}\times \Delta H& + & \text{153 g} \times 4.18 \text{ J}\cdot^{\circ}\text{C}^{-1}\cdot \text{g⁻1} \times (-8.4\, ^{\circ}\text{C}) & = & 0\\0.70\Delta H \text{ mol} & - & \text{5372 J} & = & 0\\& & 0.70\Delta H \text{ mol} & = & \text{5372 J} \\\end{array}[/tex]

[tex]\begin{array}{ccrcl} & & \Delta H & = & \dfrac{\text{ 5372 J}}{\text{0.70 mol}}\\\\ & & & = & \text{7670 J/mol}\\ & & & = & \textbf{7.7 kJ}\cdot\textbf{mol}^{ -\mathbf{-1}} \\\end{array}\\\text{The heat of solution of the compound is $\large \boxed{\textbf{7.7 kJ}\cdot\textbf{mol}^{\mathbf{-1}}}$}[/tex]

The heat of solution is positive, because the water cooled down when the substance dissolved,

To calculate ∆H for the dissolution process, we first calculate the total heat released using the equation q = m × c × ∆T and the provided properties of the solution. We then find ∆H by dividing this total heat by the number of moles of solute, yielding an enthalpy change of -7.57 kJ/mol.

The subject of this question relates to chemistry, specifically the concept of enthalpy change (∆H) in the dissolution of a solid in a solution. The first step to calculating ∆H is determining the total heat absorbed or released during the dissolution, which is calculated using the equation q = m × c × ∆T, where m is the mass of the solution, c is the specific heat capacity, and ∆T is the change in temperature.

In this case, we first convert the volume of the solution to mass by using the provided density of the solution (1.02 g/mL). So, the total mass of the solution is 150 mL × 1.02 g/mL = 153.0 g. Next, using the provided temperature and specific heat, we calculate q = (153.0 g) × (4.18 J/g･°C) × (-8.4°C) = -5300.472 J. We use a negative sign because the temperature decreases, meaning the reaction is exothermic and heat is released.

Finally, we calculate ∆H, the enthalpy change per mole of solute, by dividing q by the number of moles of the solute, 0.70 moles. ∆H = -5300.472 J / 0.70 mol = -7572 J/mol, or -7.57 kJ/mol.

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Answer:

0.098 moles of HNO₃ are present if 4.90×[tex]10^{-2}[/tex] mol of Ba(OH)₂ was needed to neutralize the acid solution

Explanation:

1 mole of Ba(OH)₂ contains 2 moles of OH- ions.

Hence, 0.049 moles of Ba(OH)2 contains x moles of OH- ions.

cross multiplying, we have

1 mole * x mole = 2 moles * 0.049 mole

x mole = 2 * 0.049 / 1

x mole =  0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present if 0.049 moles of Ba(OH)₂ was needed to neutralize the acid solution.

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

Kp = 0.0198 at 721 K for the reaction:

[tex]2HI(g)\rightleftharpoons H_2(g)+I_2(g)[/tex]

In a particular experiment, the partial pressures of H₂ and I₂ at equilibrium are 0.710 and 0.888 atm, respectively. The partial pressure of HI is __________ atm.

Answer : The partial pressure of HI is, 5.64 atm

Explanation :

For the given chemical reaction:

[tex]2HI(g)\rightleftharpoons H_2(g)+I_2(g)[/tex]

The expression of [tex]K_p[/tex] for above reaction follows:

[tex]K_p=\frac{P_{H_2}\times P_{I_2}}{(P_{HI})^2}[/tex]

We are given:

[tex]P_{H_2}=0.710atm[/tex]

[tex]P_{H_2}=0.888atm[/tex]

[tex]k_p=0.0198[/tex]

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

[tex]0.0198=\frac{0.710\times 0.888}{(P_{HI})^2}[/tex]

[tex]P_{HI}=5.64atm[/tex]

Thus, the partial pressure of HI is, 5.64 atm

The correct partial pressure of HI cannot be determined without the accurate initial conditions and changes at equilibrium, since the provided equilibrium partial pressures of H2 and I2 do not match the provided equilibrium constant.

The question states that the partial pressures of H2 and I2 at equilibrium are 0.710 atm and 0.888 atm, respectively, in a reaction H2(g) + I2(g) = 2 HI(g). Given the partial pressure at the start for HI was 1 atm and we are looking to find the equilibrium partial pressure of HI, we can apply the equilibrium constant Kp. However, the values given are part of a typo or irrelevant to the scenario as they do not match with the equilibrium constant Kp provided (1.82  imes 10-2 at 698K), and since such calculations require an accurate accounting of initial conditions and changes at equilibrium, providing an exact value for the partial pressure of HI would be speculative.

Explanation:

Answer:

CH3OH, NaBr, CaCl2, Cr(NO3)3

Based on disorderliness, the ionization strength gives the entropy.

CH3OH - Does not ionize, it's a solvent

NaBr- Ionizes intwo two ions

CaCl2- Ionizes to three ions 1Ca and 2 Cl ions

Cr(NO3)3- Ionizes to four ions

Explanation:

Here is the complete question:.

Rank the following solutes in order of increasing entropy when 0.0100 moles of each dissolve in 1.00 liter of water.

CaCl2, CH3OH, Cr(NO3)3, NaBr

To find the energy required to heat 1.00 kg of ethane from 25.0°C to 73.4°C, one must calculate the number of moles of ethane, determine the temperature change, and apply the formula Q = n • Cv • ΔT using the given specific heat capacity at constant volume (Cv).

The question pertains to the calculation of energy required for a temperature change in ethane (C2H6) and is based on thermodynamic principles involving heat, enthalpy, and specific heat capacities. To calculate the energy required to change the temperature of 1.00 kg of ethane from 25.0°C to 73.4°C in a rigid vessel, we need the specific heat capacity at constant volume (Cv) and the molar mass of ethane to convert the mass to moles. The energy, Q, needed for the temperature change is given by Q = n • Cv • ΔT, where n is the number of moles, and ΔT is the change in temperature.

First, calculate the number of moles (n) of ethane: its molar mass is 30.07 g/mol (add the atomic masses of 2 carbon atoms and 6 hydrogen atoms). Given that we have 1.00 kg (or 1000 g) of ethane, divide 1000 g by 30.07 g/mol to find n. Then, use ΔT = 73.4°C - 25.0°C to find the temperature change. Finally, plug these values into the equation Q = n • Cv • ΔT with the given Cv of 44.60 J/K·mol to calculate the required energy in joules.

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The energy required to change the temperature of 1.00 kg of ethane (C₂H₆) from 25.0°C to 73.4°C in a rigid vessel is 71622 J.

To calculate the energy required to change the temperature of 1.00 kg of ethane (C₂H₆) from 25.0°C to 73.4°C in a rigid vessel, we need to use the specific heat capacity at constant volume ([tex]C_v[/tex]) and the number of moles of ethane.

Given:

Step-by-Step Calculation

1. Convert the mass of ethane to moles:

[tex]n = \frac{m}{M}[/tex]

where

[tex]n = \frac{1000 \, \text{g}}{30.07 \, \text{g/mol}} \\\\n \approx 33.25 \, \text{mol}[/tex]

2. Calculate the temperature change:

[tex]\Delta T = T_2 - T_1 \\\\\Delta T = 73.4^\circ\text{C} - 25.0^\circ\text{C}\\\\\Delta T = 48.4 \, \text{K}[/tex]

3. Calculate the energy required using the formula:

[tex]Q = n \cdot C_v \cdot \Delta T[/tex]

where

[tex]Q = 33.25 \, \text{mol} \cdot 44.60 \, \text{J/K/mol} \cdot 48.4 \, \text{K}\\\\Q \approx 33.25 \cdot 44.60 \cdot 48.4\\\\Q \approx 71621.9 \, \text{J}[/tex]

So, the energy required to change the temperature of 1.00 kg of ethane from 25.0°C to 73.4°C in a rigid vessel is approximately:

[tex]Q \approx 71622 \, \text{J}[/tex]

The energy required is 71622 J.

Answer:

Excretory system, and Kidney.

Explanation:

The excretory system helps to regulate the chemical composition of the body fluids by removing metabolic waste such as undigested food, intestinal bacteria, and old cells and helping to retain the proper amount of water, nutrients, and salts in the body.

The main function of the kidney is to filter the blood, kidney help to remove waste from the body as urine, control fluid balance in the body, and keep the right level of electrolytes. Each kidney is bean-shaped in structure and 4-5 inches long in size.

Explanation:

More is the surface area of water comes in contact with the atmosphere or its surrounding more readily it will evaporate.

So, when 55 mL of water in a beaker with a diameter of 4.5 cm is placed then due to the small diameter of the beaker less surface area is in contact with the atmosphere.

But when 55 mL of water in a dish with a diameter of 12 cm is placed then it has larger diameter as compared to 4.5 cm diameter. Therefore, water in a beaker of 12 cm diameter will evaporate more quickly.

Thus, we can conclude that water in 55 mL of water in a dish with a diameter of 12 cm will evaporate more quickly.

The larger the surface area exposed to air, the faster the rate of evaporation. Therefore, a dish with a larger diameter will allow water to evaporate more quickly than a beaker with a smaller diameter while holding the same volume of liquid.

The 55 mL of water will evaporate more quickly in the dish with a diameter of 12 cm. This is because evaporation is a surface phenomenon - it depends on the surface area exposed to air. In this case, the dish with a more significant diameter will have a larger surface area exposed to air, leading to faster evaporation.

An analogy might be to consider two scenarios: the first is pouring a cup of coffee into a narrow, high cylinder and the second is into a broad, low dish. Both containers might have the same volume, but the surface area is significantly different. The coffee in the broad dish is more exposed and will cool faster due to a higher degree of evaporation.

Just as the coffee cools in the broad dish, the water in the dish with the larger diameter will evaporate more quickly. While volume might be constant in both the beaker and the dish, the shape and structure of the container significantly impact the rate of evaporation.

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Answer:

The molecular formula is C6H6O2

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100 grams

The compound has:

65.45 % C = 65.45 grams

5.492 % H = 5.49 grams

29.06 % O = 29.06 grams

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.00 g/mol

Molar mass = 110 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 65.45 grams / 12.01 g/mol

Moles C = 5.450 moles

Moles H = 5.49 grams / 1.01 g/mol

Moles H = 5.44 moles

Moles O = 29.06 grams / 16.00 g/mol

Moles O = 1.816 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

Moles C = 5.450 moles / 1.816 = 3

Moles H = 5.44 moles / 1.816 = 3

Moles O = 1.816/1.816 = 1

The empirical formula is C3H3O

The molar mass of this formula is 55 g/mol

We have to multiply the empirical formula by n

n = 110/55 = 2

2*(C3H3O) = C6H6O2

The molecular formula is C6H6O2

The empirical formula for the compound, given the percentage composition and molecular mass, is C3H3O. The molecular formula, given that the molar mass of the compound is about 2.66 times the empirical formula's molar mass, is C9H9O3.

To determine the molecular formula of the compound, we first assume that we have 100g of the compound. This means we have 65.45g of Carbon (C), 5.492g of Hydrogen (H), and 29.06g of Oxygen (O). We must then convert these masses to moles by dividing by the atomic mass (12.01 for C, 1.008 for H, and 16 for O).

Therefore, we have:

We then divide these mole quantities by the smallest obtained mole number to try to get whole numbers. This should give an atomic ratio of 3:3:1 for C, H, and O respectively.

By combining these ratios into a chemical formula, we get C3H3O. This is the empirical formula. The molar mass of this formula (41 g/mol) goes into the given molar mass of the compound (110 g/mol) about 2.66 times, indicating that the true molecular formula of this compound is C9H9O3.

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Answer:

A. atmospheric CO2 increases

Explanation:

The concentration of CO2 during the Paleozoic era was much higher than the current one, due to geographical factors, but not the chemical composition of the air, which should have been more important in increasing this gas.

The highest concentrations of CO2 during the Paleozoic era occurred during the Cambrian, approximately 7000 ppm (18 times higher than current measurements). During the Late Ordovic Period the CO2 concentration was approximately 12 times higher than the current ones, with a value of 4400 ppm. These high concentrations of CO2 favored the plants colonizing the continents and thus affecting the world climate.

Answer:

Current is 32.8 A

Explanation:

1 mass of Au atom = 3.27×10⁻²⁵kg

Mass of gold = 3.6 kg

We determine the amount of atoms, in that mass of gold therefore we need a rule of three:

3.27×10⁻²⁵kg is the mass for 1 atom

3.6 kg will be the mass for (3.6 . 1)/3.27×10⁻²⁵ = 1.10×10²⁵ atoms

Now we have to find out the total charge of the 3.6 kg of gold so we make another rule of three:

1 atom of Au has a charge of 1.602 × 10⁻¹⁹ C

1.10×10²⁵ atoms of Au will have a charge of (1.10×10²⁵.1.602 × 10⁻¹⁹) / 1 = 1.76×10⁶ C

Formula for the current is: q = i . t

where q is the C, i the value of current and t, time (s)- We make time conversion from h to s

14.88 h . 3600s / 1h = 53568 s  → We replace values:

1.76×10⁶ C  = i . 53568s

1.76×10⁶ C  / 53568s = i → 32.8 A

The reactions Na3PO4+CoCl2,  Na2CO3+CuCl2 and CrCl2+Li2CO3 produce precipitates.

A precipitate is said to be formed when reaction of two aqueous solutions produces an insoluble product. The insoluble solid product separates from the solution and is called a precipitate.

The full reactions are written below;

Hence, the reactions;

Produce precipitates.

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To determine if a precipitate will form or not, we need to consult solubility rules. By analyzing the compounds in each reaction, we can determine whether they contain soluble or insoluble ions. Based on this analysis, we can classify each reaction as either forming a precipitate or not.

In order to determine if a precipitate will form or not, we need to consult solubility rules. According to the solubility rules, compounds containing Li+, Na+, K+, Rb+, Cs+, and NH4+ cations, as well as compounds containing NO3- and C2H3O2- anions, are generally soluble. On the other hand, compounds containing Cl-, Br-, I-, CO32-, and most O2- anions are insoluble, except when paired with the mentioned cations. By applying these rules, we can analyze each reaction:

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Answer:

Weaker

Explanation:

The strategy here is to use Raoult´s law to calculate the theoretical vapor pressure for the concentrations given and compare it with the experimental value of 211 torr.

Raoult´s law tell us that for a binary solution

P total = partial pressure A + partial pressure B = Xa PºA + Xb PºB

where Xa and Xb are the mol fractions, and  PºA and PºB are the vapor pressures of pure A and pure B, respectively

For the solution in question we have

Ptotal = 0.312 x 55.3 torr + ( 1- 0.312 ) x 256 torr     ( XA + XB = 1 )

Ptotal = 193 torr

Since experimentally, the total vapor pressure is 211 and our theoretical value is smaller ( 193 torr ), we can conclude the interactions  solute-solvent are weaker compared to the solute-solute and solvent-solvent interactions.

Using Raoult's law, we can conclude that if the actual total vapor pressure of the solution does not match the calculated value, the solution is not ideal. Deviations from the ideal behavior indicate that the solute-solvent interactions differ in strength from solute-solute and solvent-solvent interactions.

To determine if the solution is ideal and to discuss the strengths of solute-solvent interactions compared to solute-solute and solvent-solvent interactions, we can use Raoult's law. Raoult's law states that the partial vapor pressure of each component in an ideal solution is equal to the product of the mole fraction of the component in the liquid phase and the vapor pressure of the pure component.

The expected total vapor pressure for an ideal solution can be calculated by summing the partial pressures of each component. For methanol (CH3OH), with a mole fraction of 1 - 0.312 = 0.688 and a vapor pressure of 256 torr, the expected partial pressure is partial pressure of methanol = 0.688 × 256 torr. For water (H2O), with a mole fraction of 0.312 and a vapor pressure of 55.3 torr, the expected partial pressure is partial pressure of water = 0.312 × 55.3 torr. By adding these two values together, we would get the expected total vapor pressure of an ideal solution.

If the calculated total vapor pressure using Raoult's law does not match the actual total vapor pressure of 211 torr, the solution is not ideal. In that case, the deviation indicates that the solute-solvent interactions differ in strength from the solute-solute and solvent-solvent interactions. If the actual vapor pressure is lower than expected, the solute-solvent interactions are stronger, suggesting a negative deviation. Conversely, a higher actual vapor pressure indicates weaker solute-solvent interactions relative to pure components, leading to a positive deviation.