2Al(s)+Fe2O3(s)−→−heatAl2O3(s)+2Fe(l) 2Al(s)+Fe2O3(s)→heatAl2O3(s)+2Fe(l) If 26.1 kg Al26.1 kg Al reacts with an excess of Fe2O3,Fe2O3, how many kilograms of Al2O3Al2O3 will be produced?

Answer:

C. Radiation

Explanation:

The Heat transferred through wave energy (electromagnetic waves) is Radiation.

The Heat transferred through wave energy (electromagnetic waves) is Radiation.

Answer :

(a) The number of sulfur atoms are, [tex]31.61\times 10^{23}[/tex].

(b) The mass of the mass of [tex]Fe_2(SO_4)_3[/tex] is, 1059.682 grams.

(c) The number of moles of [tex]Fe_2(SO_4)_3[/tex] is, [tex]8.63\times 10^{-3}mole[/tex]

(d) The mass of the mass of [tex]Fe_2(SO_4)_3[/tex] is, [tex]19.95\times 10^{-22}g[/tex]

Explanation :

(a) As we are given the number of moles of [tex]Fe_2(SO_4)_3[/tex] is, 1.75 mole. Now we have to calculate the number of sulfur atoms.

In the [tex]Fe_2(SO_4)_3[/tex], there are 2 iron atoms, 3 sulfur atoms, 12 oxygen atoms.

As, 1 mole of [tex]Fe_2(SO_4)_3[/tex] contains [tex]3\times 6.022\times 10^{23}[/tex] number of sulfur atoms.

So, 1.75 mole of [tex]Fe_2(SO_4)_3[/tex] contains [tex]1.75\times 3\times 6.022\times 10^{23}=31.61\times 10^{23}[/tex] number of sulfur atoms.

The number of sulfur atoms are, [tex]31.61\times 10^{23}[/tex]

(b) As we are given the number of moles of [tex]Fe_2(SO_4)_3[/tex] is, 2.65 mole. Now we have to calculate the mass of [tex]Fe_2(SO_4)_3[/tex].

[tex]\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3[/tex]

The molar mass of [tex]Fe_2(SO_4)_3[/tex] = 399.88 g/mole

[tex]\text{Mass of }Fe_2(SO_4)_3=2.65mole\times 399.88g/mole=1059.682g[/tex]

The mass of the mass of [tex]Fe_2(SO_4)_3[/tex] is, 1059.682 grams.

(c) As we are given the mass of [tex]Fe_2(SO_4)_3[/tex] is, 3.45 grams. Now we have to calculate the moles of [tex]Fe_2(SO_4)_3[/tex].

[tex]\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3[/tex]

The molar mass of [tex]Fe_2(SO_4)_3[/tex] = 399.88 g/mole

[tex]3.45g=\text{Moles of }Fe_2(SO_4)_3\times 399.88g/mole[/tex]

[tex]\text{Moles of }Fe_2(SO_4)_3=8.63\times 10^{-3}mole[/tex]

The number of moles of [tex]Fe_2(SO_4)_3[/tex] is, [tex]8.63\times 10^{-3}mole[/tex]

(d) As we are given the formula unit of [tex]Fe_2(SO_4)_3[/tex] is, 3. Now we have to calculate the mass of [tex]Fe_2(SO_4)_3[/tex].

As we know that 1 mole of [tex]Fe_2(SO_4)_3[/tex] contains [tex]6.022\times 10^{23}[/tex] formula unit.

Formula used :

[tex]\text{Formula unit of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}[/tex]

[tex]3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}[/tex]

[tex]\text{Moles of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole[/tex]

Now we have to calculate the mass of [tex]Fe_2(SO_4)_3[/tex].

[tex]\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3[/tex]

The molar mass of [tex]Fe_2(SO_4)_3[/tex] = 399.88 g/mole

[tex]\text{Mass of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole\times 399.88g/mole=19.95\times 10^{-22}g[/tex]

The mass of the mass of [tex]Fe_2(SO_4)_3[/tex] is, [tex]19.95\times 10^{-22}g[/tex]

Final answer:

The key parts of solving chemistry problems include understanding chemical formulas and conducting mole-mass calculations. Using the formula for ferric sulfate, Fe(SO₄)₃, this response walks through how to find the number of atoms, mass of substance, and moles given specific quantities, showcasing the essential chemical calculations involved.

Explanation:

The chemical formula for ferric sulfate is Fe(SO₄)₃. Let's solve each part of the question:

a) The number of sulfur atoms in 1.75 mole of Fe(SO₄)₃: There are 3 sulfur atoms in one formula unit of Fe(SO₄)₃. Therefore, in 1.75 moles of Fe(SO₄)₃, there are 1.75 moles * 3 sulfur atoms per mole = 5.25 moles of sulfur atoms.

b) The mass in grams of 2.65 mol of Fe(SO₄)₃: First, we need to calculate the molar mass of Fe(SO₄)₃ (Fe = 55.85, S = 32.06, O = 16.00). The molar mass of Fe(SO₄)₃ is 399.88 g/mol. Thus, 2.65 moles * 399.88 g/mol = 1059.68 grams of Fe(SO₄)₃.

c) The number of moles of Fe(SO₄)₃ in 3.45 grams of Fe(SO₄)₃: Using the molar mass of Fe(SO₄)₃, 3.45 g / 399.88 g/mol = 0.00863 moles of Fe(SO₄)₃.

d) The mass in grams of 3 formula unit of Fe(SO₄)₃: The mass of one mole of Fe(SO₄)₃ is 399.88 g. Since one mole contains Avogadro's number (6.022 x 10²³) of formula units, 3 formula units' mass can be calculated as (3 / 6.022 x 10²³) * 399.88 g, which equals approximately 1.99 x 10⁻²² grams of Fe(SO₄)₃.

Answer: All the given options have the same number of particles.

Explanation:

For the given options:

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.

If, 22.99 g of sodium metal contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, 22.990 g of sodium metal will contain [tex]\frac{6.022\times 10^{23}}{22.99}\times 22.990=6.022\times 10^{23}[/tex] number of atoms.

Number of particles = [tex]6.022\times 10^{23}[/tex]

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, 1 mole of sodium element will also contain [tex]6.022\times 10^{23}[/tex] number of atoms.

Number of particles = [tex]6.022\times 10^{23}[/tex]

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.

If, 9.01 g of beryllium metal contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, 9.012 g of beryllium metal will contain [tex]\frac{6.022\times 10^{23}}{9.01}\times 9.012=6.022\times 10^{23}[/tex] number of atoms.

Number of particles = [tex]6.022\times 10^{23}[/tex]

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, 1 mole of beryllium element will also contain [tex]6.022\times 10^{23}[/tex] number of atoms.

Number of particles = [tex]6.022\times 10^{23}[/tex]

Hence, all the options have same number of particles which is [tex]6.022\times 10^{23}[/tex]

Both 1 mole of sodium (Na) and 1 mole of beryllium (Be) contain the greatest number of particles, which is Avogadro's Number or 6.02×1023 particles. The other listed quantities, 22.990 g of Na and 9.012 g of Be, contain fewer particles as they are less than one mole of their respective elements.

To determine which of the given options has the greatest number of particles, we must refer to Avogadro's Number and the concept of molar mass. Avogadro's Number, which is 6.02×1023 mol-1, represents the number of particles in one mole of a substance. The molar mass is the mass of one mole of representative particles of a substance and is equivalent numerically to the atomic mass of an element in grams. For instance, the atomic mass of sodium (Na) is approximately 23 g/mol, which means that 1 mole of Na has a mass of approximately 23 grams and contains 6.02×1023 particles of Na.

Therefore, 1 mole of any substance, whether Na or beryllium (Be), will always contain 6.02×1023 particles. Comparing the options given:

Both 1 mole of Na and 1 mole of Be have the greatest number of particles, which is Avogadro's Number of particles. The quantities that are less than one mole have fewer particles in comparison.

Hey there!:

1 in. Hg = 0.4911 psi

=> Absolute pressure in psi = 22.4 + 28.6 x 0.4911 = 36.445 psi

a) 1 psi = 144 lb/ft²

36.445 psi = 5248.15 lb/ft²

b) 1 psi = 2.036 in. Hg

36.445 psi = 74.21 in. Hg

Hope this helps!

1 in. Hg = 0.4911 psi

Absolute pressure in psi = 22.4 + 28.6 x 0.4911 = 36.445 psi

a) 1 psi = 144 lb/ft²

36.445 psi = 5248.15 lb/ft²

b) 1 psi = 2.036 in. Hg

36.445 psi = 74.21 in. Hg

Pressure plays a important role in rate of reaction to increase the pressure it increases the rate of reaction by increasing the collision.Pressure increase the concentration of gases it means that it increases the number of molecules per unit volume due to this collision of gases increases this will increase the temperature.

This increase of temperature will increase the rate of reaction. So we can say that the increases of pressure will increase the rate of reaction. Density is defined as the mass per unit volume it means that mass present in 1 meter cube is called density. The S.I unit of density is kg/m^3 and in C.G.S it is gram/cm^3

So In above statement we can understand that density, mass, and volume all are convert to each other it means that if we know any two variable then third one will be calculated easily.

Therefore, a) 1 psi = 144 lb/ft²

36.445 psi = 5248.15 lb/ft²

b) 1 psi = 2.036 in. Hg

36.445 psi = 74.21 in. Hg

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To calculate the equilibrium concentrations of reactant and products, we use the equilibrium constant (Kc) and the initial concentrations of the reactants and products. Using the stoichiometry of the reaction, we can determine the change in concentrations. Solve for x in the equilibrium constant expression to find the equilibrium concentrations.

To calculate the equilibrium concentrations of reactant and products, we need to use the equilibrium constant (Kc) and the initial concentrations of the reactants and products. In this case, we have 0.442 moles of CO and 0.442 moles of Cl2 in a 1.00 L vessel at 600 K. Using the stoichiometry of the reaction, we can determine the change in concentrations. Let's denote the change in concentration of COCl2 as x.

The initial concentration of CO is [CO] = 0.442 M, so the change in concentration is -x (due to the decrease in CO).

The initial concentration of Cl2 is [Cl2] = 0.442 M, so the change in concentration is -x (due to the decrease in Cl2).

The initial concentration of COCl2 is [COCl2] = 0 M, so the change in concentration is +x (due to the formation of COCl2).

Using the equilibrium constant expression Kc = [COCl2] / ([CO][Cl2]), we can set up the following equation: 77.5 = x / (0.442 * 0.442)

Solving for x, we get x = 0.442 * 0.442 * 77.5 = 15.558 M. Therefore, the equilibrium concentrations are: [CO] = 0.442 - 15.558 = -15.116 M (neglecting the negative sign), [Cl2] = 0.442 - 15.558 = -15.116 M (neglecting the negative sign), and [COCl2] = 15.558 M.

Answer:

50.8g

Explanation:

Given parameters:

Nubmer of mole of Cu₂O = 0.4mol

Unknown:

Mass of Cu produced = ?

Solution:

The balanced reaction equation:

           Cu₂O  +  H₂   ⇄ 2Cu + H₂O

From the reaction, we know that:

            1 mole of Cu₂O produced 2 moles of Cu

             0.4 mol of Cu₂O would give 0.8mol of Cu

Now we know the number of moles of Cu produced because the limiting reactant here is Cu₂O. It was used up in the reaction in which the hydrogen gas was in excess.

To find the mass of Cu produced, we use the equation below :

              Mass of Cu = Number of moles of Cu x molar mass of Cu

Molar mass of Cu = 63.5gmol⁻¹

              Mass of Cu = 0.8 x 63.5 = 50.8g

If [tex]0.40[/tex] mol of [tex]Cu_2O(s)[/tex] is reduced completely with excess [tex]H_2(g)[/tex], approximately [tex]50.84[/tex] grams of [tex]Cu(s)[/tex] would be produced.

The balanced chemical equation for the reduction of [tex]Cu_2O(s)[/tex] with [tex]H_2(g)[/tex] is:

[tex]\[ \text{Cu}_2\text{O}(s) + \text{H}_2(g) \rightarrow \text{Cu}(s) + \text{H}_2\text{O}(l) \][/tex]

From the balanced equation, we can see that [tex]1[/tex] mole of [tex]Cu_2O[/tex] produces [tex]2[/tex] moles of [tex]Cu[/tex].

Given that we have [tex]0.40[/tex] mol of [tex]Cu_2O[/tex], we can use the stoichiometry of the reaction to find the amount of [tex]Cu[/tex] produced:

[tex]\[ \text{mol of Cu} = 0.40 \, \text{mol Cu}_2\text{O} \times \frac{2 \, \text{mol Cu}}{1 \, \text{mol Cu}_2\text{O}} \][/tex]

[tex]\[ \text{mol of Cu} = 0.40 \times 2 \][/tex]

[tex]\[ \text{mol of Cu} = 0.80 \, \text{mol} \][/tex]

Now, we need to convert the moles of [tex]Cu[/tex] to grams using the molar mass of copper ([tex]Cu[/tex]), which is approximately [tex]63.55 g/mol[/tex]

[tex]\[ \text{mass of Cu} = \text{mol of Cu} \times \text{molar mass of Cu} \][/tex]

[tex]\[ \text{mass of Cu} = 0.80 \, \text{mol} \times 63.55 \, \text{g/mol} \][/tex]

[tex]\[ \text{mass of Cu} = 50.84 \, \text{g} \][/tex]

Answer: 0.736 g

Explanation:

[tex]O_3+NO\rightarrow O_2+NO2[/tex]

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

[tex]\text{Number of moles of}O_3=\frac{0.781g}{48g/mol}=0.016moles[/tex]

[tex]\text{Number of moles of}NO=\frac{0.589g}{30g/mol}=0.019moles[/tex]

By Stoichiometry:

1 mole of ozone [tex]O_3[/tex] reacts with 1 mole of nitric oxide [tex]NO[/tex] to form 1 mole of nitrogen dioxide [tex]NO_2[/tex]

0.016 moles of ozone reacts with=[tex]\frac{1}{1}\times 0.016=0.016moles[/tex] of nitric oxide to form 0.016 mole of [tex]NO_2[/tex]

Thus ozone is the limiting reagent as it limits the formation of products and nitric oxide is the excess reagent as (0.019-0.016) g= 0.003 g remains as such.

Mass of [tex]NO_2=moles\times {\text{Molar mass}}=0.016\times 46=0.736g[/tex]

0.736 g of [tex]NO_2[/tex] will be produced.

0.748 grams of [tex]\( \text{NO}_2 \)[/tex] will be produced when 0.781 grams of [tex]\( \text{O}_3 \)[/tex] reacts with 0.589 grams of NO, as [tex]\( \text{O}_3 \)[/tex] is the limiting reactant.

To determine how many grams of [tex]\( \text{NO}_2 \)[/tex] will be produced from the reaction of [tex]\( \text{O}_3 \)[/tex] with NO, we need to follow these steps:

1. Write the balanced chemical equation:

[tex]\text{O}_3 + \text{NO} \longrightarrow \text{O}_2 + \text{NO}_2[/tex]

2. Calculate the moles of reactants:

Moles of [tex]\( \text{O}_3 \)[/tex]:

[tex]\text{Molar mass of } \text{O}_3 = 3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g/mol}[/tex]

[tex]\text{Moles of } \text{O}_3 = \frac{0.781 \, \text{g}}{48.00 \, \text{g/mol}} = 0.01627 \, \text{mol}[/tex]

Moles of NO:

[tex]\text{Molar mass of } \text{NO} = 14.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 30.01 \, \text{g/mol}[/tex]

[tex]\text{Moles of } \text{NO} = \frac{0.589 \, \text{g}}{30.01 \, \text{g/mol}} = 0.01963 \, \text{mol}[/tex]

3. Determine the limiting reactant:

The balanced equation shows that 1 mole of [tex]\( \text{O}_3 \)[/tex] reacts with 1 mole of NO.

Compare the moles of each reactant:

[tex]0.01627 \, \text{mol} \, \text{O}_3 \quad \text{and} \quad 0.01963 \, \text{mol} \, \text{NO}[/tex]

Since [tex]\( \text{O}_3 \)[/tex] has fewer moles, it is the limiting reactant.

4. Calculate the moles of [tex]\( \text{NO}_2 \)[/tex] produced:

From the balanced equation, 1 mole of [tex]\( \text{O}_3 \)[/tex] produces 1 mole of [tex]\( \text{NO}_2 \)[/tex].

Therefore, moles of [tex]\( \text{NO}_2 \)[/tex] produced = moles of [tex]\( \text{O}_3 \)[/tex] reacted:

[tex]\text{Moles of } \text{NO}_2 = 0.01627 \, \text{mol}[/tex]

5. Convert moles of [tex]\( \text{NO}_2 \)[/tex] to grams:

Molar mass of [tex]\( \text{NO}_2 \)[/tex]:

[tex]\text{Molar mass of } \text{NO}_2 = 14.01 \, \text{g/mol} + 2 \times 16.00 \, \text{g/mol} = 46.01 \, \text{g/mol}[/tex]

Grams of [tex]\( \text{NO}_2 \)[/tex]:

[tex]\text{Mass of } \text{NO}_2 = 0.01627 \, \text{mol} \times 46.01 \, \text{g/mol} = 0.748 \, \text{g}[/tex]

When 4.11 g of citric acid reacts with sodium bicarbonate, 2.82 g of carbon dioxide gas is produced according to the provided chemical reaction.

In the given chemical reaction, 3NaHCO3 + C6H8O7 → 3CO2 + 3H2O + Na3C6H5O7, specifically between sodium bicarbonate and citric acid, it's clear that for every one molecule of citric acid, three molecules of carbon dioxide are formed. Given that you have 4.11 g of citric acid, and knowing that the molecular weight of citric acid is 192 g/mol, first we calculate the number of moles of citric acid by dividing the mass by the molecular weight: 4.11 g/192 g/mol = 0.0214 moles. Since each mole of citric acid react to produce 3 moles of carbon dioxide, then 0.0214 moles of citric acid would produce 0.0214 moles x 3 = 0.0642 moles of CO2. To find the mass of CO2 formed (since its molar mass is 44 g/mol), multiply the resulting moles of CO2 by its molar mass: 0.0642 moles x 44 g/mol = 2.82 g. Therefore, when 4.11 g of citric acid reacts with sodium bicarbonate, 2.82 g of carbon dioxide will form.

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Answer:

3

Explanation:

Lt= Loe^(-kt)

Data:

Lo = 10 mg/mL

Assume k = 0.23/da

1. Calculate L5

L5 = 10e^(-5×0.23) = 10e^-1.15 = 10 × 0.317 = 3.17 mg/mL

2. Calculate the dilution factor

You expect to find L5 to be about 3

You want L5 to be about 1.

You should use a dilute your sample by a factor of 3.

P = 3

Answer:

The rate at which dihydrogen is being produced is 0.12 kg/sec.

Explanation:

[tex]CH_4+H_2O\rightarrow CO+3H_2[/tex] Haber reaction

Volume of methane consumed in a second = 924 L

Temperature at which reaction is carried out,T= 261°C = 538.15 K

Pressure at which reaction is carried out, P = 0.96 atm

Let the moles of methane be n.

Using an Ideal gas equation:

[tex]PV=nRT[/tex]

[tex]n=\frac{PV}{RT}=\frac{0.96 atm\times 924 L}{0.0821 atm l/mol K\times 538.15 K}=20.0769 mol[/tex]

According to reaction , 1 mol of methane gas produces 3 moles of dihydrogen gas.

Then 20.0769 mol of dihydrogen will produce :

[tex]\frac{3}{1}\times 20.0769 mol=60.2307 mol[/tex] of dihydrogen

Mass of 24.3194 moles of ammonia =24.3194 mol × 2 g/mol

=120.46 g=0.12046 kg ≈ 0.12 kg

924 L of methane are consumed in 1 second to produce 0.12 kg of dihydrogen in 1 second. So the rate at which dihydrogen is being produced is 0.12 kg/sec.

Answer:

Increasing order of boiling point

isopropyl alcohol > acetone > pentane > ethane

Explanation:

Vapor pressure - the pressure exerted the gaseous molecules , on the walls of the container is called the vapor pressure.

Boiling point - the temperature at which the the vapor pressure of the liquid equals the external atmospheric pressure.

Both boiling point and vapor pressure are linked by the inter molecular forces between the atoms.

The compound with stronger inter molecular forces are tightly held , hence more amount of energy is required to vaporize. Therefore, higher boiling point , and in turn the vapor pressure will be lower .

And the compound with weaker inter molecular forces are loosely held , hence less amount of energy is required to vaporize. Therefore, lower boiling point , and in turn the vapor pressure will be higher .

Therefore,

The vapor pressure and boiling point will have an inverse relation.

Given , The vapor pressure of the compound in increasing order -

isopropyl alcohol < acetone < pentane < ethane .

Hence, the order of boiling point will be exactly reverse.

So , the Increasing order of boiling point,

isopropyl alcohol > acetone > pentane > ethane

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

Answer : The total change of volume is, 41.883 liters.

Explanation :

R134a is a 1,1,1,2-tetrafluoroethane. It is a hydro-fluorocarbon and haloalkane gaseous refrigerant.

First we have to calculate the volume at [tex]-20^oC[/tex].

Using ideal gas equation:

[tex]PV=nRT\\\\PV=\frac{w}{M}\times RT[/tex]

where,

n = number of moles

w = mass of R134a = 257 g

P = pressure of the gas = 60 Kpa

T = temperature of the gas = [tex]-20^oC=273+(-20)=253K[/tex]

M = molar mass of R134a  = 102.03 g/mole

R = gas constant = 8.314 Kpa.L/mole.K

V = initial volume of gas

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

[tex](60Kpa)\times V=\frac{257g}{102.03g/mole}\times (8.314Kpa.L/mole.K)\times (253K)[/tex]

[tex]V=88.305L[/tex]

Now we have to calculate the volume at [tex]100^oC[/tex] by using Charles's law.

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

[tex]V\propto T[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 88.305 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = 253 K

[tex]T_2[/tex] = final temperature of gas = [tex]100^oC=273+100=373K[/tex]

Now put all the given values in the above formula, we get the final volume of the gas.

[tex]\frac{88.305L}{V_2}=\frac{253K}{373K}[/tex]

[tex]V_2=130.188L[/tex]

Now we have to calculate the total change of volume.

[tex]V_2-V_1=130.188-88.305=41.883L[/tex]

Therefore, the total change of volume is, 41.883 liters.

The mass of sodium phosphate needed to make 375 mL of a solution with a molarity of sodium ions of 0.900 M is indeed 18.45 grams.

Given:

Volume of solution (V) = 375 mL = 0.375 L

Molarity of sodium ions (c) = 0.900 M

Using the formula for molarity: c = n/V

Since each mole of sodium phosphate produces 3 moles of sodium ions:

Effective molarity of sodium phosphate solution = c / 3 = 0.300 M

Number of moles of sodium phosphate (n) = c * V = 0.300 M * 0.375 L = 0.1125 mol

Molar mass of sodium phosphate (M) = 164 g/mol

Mass of sodium phosphate (m) = n * M = 0.1125 mol * 164 g/mol = 18.45 g

Therefore, the mass of sodium phosphate needed to make 375 mL of a solution with a molarity of sodium ions of 0.900 M is indeed 18.45 grams.

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To make a solution of tribasic sodium phosphate with a concentration of 1.50 M Na+ ions in 375 mL, you will need 276.38 grams of Na3PO4.

To calculate the grams of Na3PO4 needed, we need to use the equation relating moles, volume, and concentration:

Moles of Na+ ions = Molarity of Na+ ions x Volume of solution (in L)

Moles of Na3PO4 = 3 x Moles of Na+ ions

First, convert the volume in mL to L (375 mL = 0.375 L). Then, substitute the values into the equation:

Moles of Na+ ions = 1.50 M x 0.375 L = 0.5625 mol

Moles of Na3PO4 = 3 x 0.5625 mol = 1.6875 mol

Finally, convert moles to grams using the molar mass of Na3PO4 (163.94 g/mol):

Mass of Na3PO4 = 1.6875 mol x 163.94 g/mol = 276.38 g

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Determine the standard cell potential for the reaction from given standard reduction potentials. Use this in the Nernst equation to calculate the equilibrium constant (Keq). From Keq, interpret the system at equilibrium.

The calculation involved in this chemistry question is related to electrochemistry, specifically determining the equilibrium constant (Keq) for a redox reaction involving copper and cobalt in an electrochemical cell. This calculation can be performed using standard reduction potentials and the Nernst equation.

Based on the given half-reactions, the net reaction is Cu2+ (aq) + Co(s) → Cu(s) + Co2+ (aq). You would first determine the E°cell (standard cell potential) for the reaction by subtracting the standard reduction potential for the oxidation half-reaction (occurring at the anode) from that for the reduction half-reaction (at the cathode).

Assuming we have the standard reduction potential values for the cathode and anode processes, E.cell can be calculated. Note that at equilibrium, E.cell equals to 0, so from the Nernst equation we derive the formula for Keq. Remember that n is the number of moles of electrons transferred in the reaction, which is equivalent to 2 in this reaction. From this calculated Keq value, you will be able to infer changes in the system at equilibrium.

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Answer: 1. [tex]2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O[/tex]

2. 100.8 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Balanced equation for the decomposition of the compound sodium hydrogen carbonate is:

[tex]2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O[/tex]

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

moles of [tex]CO_2=\frac{\text {given mass}}{\text {molar mass}}=\frac{27.0g}{44g/mol}=0.6moles[/tex]

According to stoichiometry:

1 mole of [tex]CO_2[/tex] is produced from 2 moles of [tex]NaHCO_3[/tex]

Thus 0.6 moles of [tex]CO_2[/tex] is produced from =[tex]\frac{2}{1}\times 0.6=1.2[/tex] moles of [tex]NaHCO_3[/tex]

Mass of  [tex]NaHCO_3=moles\times {\text {Molar mass}}=1.2\times 84=100.8g[/tex]

Thus  the mass of [tex]NaHCO_3[/tex] required to produce 27.0 g of  [tex]CO_2[/tex] is 100.78 grams.

The balanced equation for the decomposition of sodium bicarbonate is 2 NaHCO3 → Na2CO3 + CO2 + H2O. To produce 27.0 g of CO2, you would need approximately 51.0 g of NaHCO3.

The balanced chemical equation for the decomposition of sodium bicarbonate or baking soda is 2 NaHCO3 → Na2CO3 + CO2 + H2O. When heated, sodium bicarbonate decomposes into sodium carbonate, carbon dioxide, and water. This CO2 is what makes baked goods rise.

To calculate the mass of NaHCO3 required to produce 27.0 g of CO2, we use molar mass and stoichiometry. The molar mass of NaHCO3 is about 84 g/mol, and that of CO2 is about 44 g/mol. Since the reaction produces one mole of CO2 for every two moles of NaHCO3, we can set up the equation: (27g CO2 / 44 g/mol CO2) * (2 mol NaHCO3 / 1 mol CO2) * (84 g NaHCO3 / 1 mol NaHCO3) = 51.0 g NaHCO3

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Answer:

The final concentration is 0.2497 M.

Explanation:

1. Molarity of the starting solution = [tex]M_1=1.80 M[/tex]

Volume of the starting solution = [tex]V_1=60.0 mL[/tex]

After dilution of the solution to 219 mL.

Molarity of solution after dilution = [tex]M_2[/tex]

Volume of the new solution = [tex]V_2=218 mL[/tex]

[tex]M_1\times V_1=M_2\times V_2[/tex] (Dilution law)

[tex]M_2=\frac{1.80 M\times 60.0 mL}{218 mL}=0.495 M[/tex]

2. Now,109 mL of 0.495 M solution was diluted by 107 mL

Molarity of the solution = [tex]M_1=0.495 M[/tex]

Volume of the solution = [tex]V_1=109 mL[/tex]

Molarity of solution after dilution = [tex]M_2[/tex]

Volume of the new solution(107 mL of water is added)  = [tex]V_2=109 mL+107 mL = 216 mL[/tex]

[tex]M_1\times V_1=M_2\times V_2[/tex] (Dilution law)

[tex]M_2=\frac{0.495 M\times 109 mL}{216 mL}=0.2497 M[/tex]

The final concentration is 0.2497 M.

To calculate the final concentration after multiple dilutions, use the dilution formula (M1V1 = M2V2) for each dilution step. The first dilution involves diluting a 60.0 mL sample of a 1.80 M solution to 218 mL. The second dilution involves diluting a 109 mL sample of this diluted solution by adding 107 mL of water.

The calculation involved here relies on the formula used for dilution which states that M1V1 = M2V2, where M represents molarity and V represents volume. Initially, we have a 1.80 M solution and we take a 60.0 mL aliquot. After dilution, this becomes a 218 mL solution. Hence, the new molarity (M2) can be calculated as (1.80 M * 60.0 mL) / 218 mL.

From that diluted solution, we take a 109 mL sample and add 107mL of water to it, again diluting this solution further. This process is again similar to the first step of dilution. To find the final concentration, use the dilution formula again with the newly calculated molarity as M1 and 109 mL as V1. The final volume (V2) would be 109 mL + 107 mL. So the final concentration will be given by: (M1 * 109 mL) / (109 mL + 107 mL).

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Answer:

0.544 eV; Ag is a smaller atom.

Explanation:

1. Ionization energy of hydrogen

The outermost electrons in Rb and Ag are in 5s orbitals.

The formula for the energy of an electron in a hydrogen atom is

E = -13.6/n² eV

For a hydrogen atom in a 5s orbital,

E = -13.6/5² = -13.6/25 = -0.544 eV

The ionization energy would be 0.544 eV.

2. Rb vs Ag

The first electrons to be removed from Rb and Ag are in 5s orbitals.

The atomic radius of Ag is less than that of Rb because, as you go from left to right across the Row, you are adding 10 protons to the nucleus and 10 electrons to the outer shell.

The added electrons do not effectively shield each other from the attraction of the nucleus, so the 5s electron of Ag is closer in.

It takes more energy to remove the electron from silver, so the ionization energy of Ag is greater than that of Rb.

Rubidium's and silver's higher ionization energies arise from effective nuclear charge, quantum defect, and relativistic effects.

The ionization energies of rubidium and silver are 4.18 and 7.57 eV, respectively. To calculate the ionization energies of an H atom with its electron in the same outermost orbitals as in these two atoms, we must consider the effective nuclear charge experienced by the outermost electron in each atom.

For hydrogen, the ionization energy is given by the formula:

[tex]\[ IE = \frac{13.6 \text{ eV}}{n^2} \][/tex]

where n is the principal quantum number of the electron's orbital.

For rubidium (Rb), the outermost electron is in the 5s orbital, which corresponds to n = 5. Therefore, the ionization energy for a hydrogen-like atom with an electron in the 5s orbital is:

[tex]\[ IE_{\text{H, Rb-like}} = \frac{13.6 \text{ eV}}{5^2} = \frac{13.6 \text{ eV}}{25} \approx 0.544 \text{ eV} \][/tex]

For silver (Ag), the outermost electron is in the 5s orbital as well, which also corresponds to n = 5. Thus, the ionization energy for a hydrogen-like atom with an electron in the 5s orbital is the same as for the rubidium-like case:

[tex]\[ IE_{\text{H, Ag-like}} = \frac{13.6 \text{ eV}}{5^2} = \frac{13.6 \text{ eV}}{25} \approx 0.544 \text{ eV} \][/tex]

However, the actual ionization energies of rubidium and silver are much higher than the calculated values for hydrogen-like atoms because of the following reasons:

1. Penetration Effect:

The inner electrons shield the outermost electron from the nucleus. In a multi-electron atom, the effective nuclear charge [tex](\( Z_{\text{eff}} \))[/tex] is less than the actual nuclear charge due to this shielding.

The outermost electron in rubidium and silver experiences a [tex]\( Z_{\text{eff}} \)[/tex] that is less than the full nuclear charge.

2. Quantum Defect:

The quantum defect accounts for the difference in energy levels between hydrogen-like atoms and many-electron atoms due to the presence of inner electrons.

This defect is more significant for atoms with higher atomic numbers, like silver.

3. Relativistic Effects:

For heavier atoms like silver, relativistic effects become significant.

The mass of the electron increases as it moves at a significant fraction of the speed of light, which increases the electron's binding energy to the nucleus, thus increasing the ionization energy.

4. Full d Subshell:

Silver has a full 4d subshell, which provides additional stability and pulls the 5s orbital closer to the nucleus, increasing the ionization energy.

Due to these effects, the actual ionization energies of rubidium and silver are much higher than those calculated for a hydrogen-like atom.

The differences in values between rubidium and silver are primarily due to the increased nuclear charge, the presence of filled d subshells in silver, and relativistic effects, which are more pronounced in silver due to its higher atomic number.

The molal boiling point elevation constant, Kb, of the liquid X in the presence of the given amount of urea is calculated to be approximately 2.239 °C kg/mol.

The question requires us to calculate the molal boiling point elevation constant, Kb, of a fluid X. The molal boiling point elevation constant is a measure of how much the boiling point of a solution increases when a solute is added. This is represented by the formula: ∆TB = Kb*m, where ∆TB is the change in boiling point and m is the molal concentration of the solute. Given that

∆TB = TB(solution) - TB(pure solvent) = 123.0°C - 120.7°C = 2.3°C,

and the molality m = moles of solute/kg of solvent = (58.66g urea / 60.056 g/mol urea) / (950g solvent / 1000g/kg) = 1.027 mol/kg,

The molal boiling point elevation constant, Kb, can be calculated by rearranging ∆TB = Kb*m to Kb = ∆TB / m = 2.3°C / 1.027 mol/kg = 2.239 °C kg/mol.

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Based on boiling point elevation, mole number, and molality, the molal boiling point elevation constant (Kb) of a liquid X containing 58.66g of urea is computed as 2.23 °C/m.

Step-by-Step Solution:

Determine the boiling point elevation (∆Tb):

[tex]\Delta T_b = 123.0\textdegree C - 120.7\textdegree C = 2.3\textdegree C[/tex]

Calculate the number of moles of urea (NH₂₂CO):

[tex]Molar\ mass\ of\ urea\ = 14 + 1\times 4 + 12 + 16 = 60\ g/mol[/tex]

Urea moles are equal to 58.66 g / 60 g/mol, or 0.978 moles.

Calculate the molality (m) of the solution:

Molality (m) is calculated as solute moles per kilogram of solvent.

[tex]m = 0.978\ moles / 0.950\ kg = 1.03\ m[/tex]

Use the boiling point elevation formula to calculate Kb:

[tex]\Delta T_b = K_b \times m[/tex]

[tex]K_b = \Delta T_b / m = 2.3\textdegree C / 1.03\ m = 2.23 \textdegree C/m[/tex]

As a result, liquid X's molal boiling point elevation constant, or Kb, is 2.23 °C/m.

Answer : The mass of silver chloride formed are, 8.44 grams.

Explanation : Given,

Mass of [tex]AgNO_3[/tex] = 10.0 g

Mass of [tex]BaCl_2[/tex] = 15.0 g

Molar mass of [tex]AgNO_3[/tex] = 169.87 g/mole

Molar mass of [tex]BaCl_2[/tex] = 208.23 g/mole

Molar mass of [tex]AgCl[/tex] = 143.32 g/mole

First we have to calculate the moles of [tex]AgNO_3[/tex] and [tex]BaCl_2[/tex].

[tex]\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{10g}{169.87g/mole}=0.0589moles[/tex]

[tex]\text{Moles of }BaCl_2=\frac{\text{Mass of }BaCl_2}{\text{Molar mass of }BaCl_2}=\frac{15g}{208.23g/mole}=0.072moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]2AgNO_3+BaCl_2\rightarrow 2AgCl+Ba(NO_3)2[/tex]

From the balanced reaction we conclude that

As, 2 moles of [tex]AgNO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]

So, 0.0589 moles of [tex]AgNO_3[/tex] react with [tex]\frac{0.0859}{2}=0.0295[/tex] moles of [tex]BaCl_2[/tex]

From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]AgCl[/tex].

As, 2 moles of [tex]AgNO_3[/tex] react to give 2 moles of [tex]AgCl[/tex]

So, 0.059 moles of [tex]AgNO_3[/tex] react to give 0.059 moles of [tex]AgCl[/tex]

Now we have to calculate the mass of [tex]AgCl[/tex].

[tex]\text{Mass of }AgCl=\text{Moles of }AgCl\times \text{Molar mass of }AgCl[/tex]

[tex]\text{Mass of }AgCl=(0.0589mole)\times (143.32g/mole)=8.44g[/tex]

Therefore, the mass of silver chloride formed are, 8.44 grams.

The mass of silver chloride (AgCl) formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride is 8.44 g. The correct option is the second option 8.44 g

First, we will write a balanced chemical equation for the reaction between silver nitrate and barium chloride

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

This means, 2 moles of AgNO₃ will react with 1 mole of BaCl₂ to produce 2 moles of AgCl and 1 mole of Ba(NO₃)₂

From the question,

10.0 g of silver nitrate reacts with 15.0 g of barium chloride

First, we will determine the number of moles present in each reactant

Mass = 10.0 g

Molar mass = 169.87 g/mol

Using the formula

[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]

Then,

[tex]Number \ of\ moles\ of\ silver\ nitrate = \frac{10.0}{169.87}[/tex]

Number of moles of AgNO₃ = 0.05887 moles

Mass = 15.0 g

Molar mass of BaCl₂ = 208.23 g/mol

∴ [tex]Number \ of\ moles\ of\ barium \ chloride= \frac{15.0}{208.23}[/tex]

Number of moles of BaCl₂ = 0.072036 moles

From the balanced chemical equation

2 moles of AgNO₃ will react with 1 mole of BaCl₂

∴ 0.05887 moles AgNO₃ will react with [tex]\frac{0.05887}{2}[/tex] mole of BaCl₂

[tex]\frac{0.05887}{2} = 0.029435[/tex]

This means only 0.029435 moles of BaCl₂ will react

(NOTE: AgNO₃ is the limiting reagent while BaCl₂ is the excess reagent)

Now,

Since 2 moles of AgNO₃ will react with 1 mole of BaCl₂ to produce 2 moles of AgCl

That means,

0.05887 moles AgNO₃ will react with 0.029435 moles of BaCl₂ to produce 0.05887 moles of AgCl

∴ The number of moles of silver chloride (AgCl) produced is 0.05887 moles

Now, to determine the mass (in grams) of silver chloride that are formed

From the formula,

Mass = Number of moles × Molar mass

Molar mass of AgCl = 143.32 g/mol

∴ Mass of silver chloride, AgCl, formed = 0.05887 moles × 143.32 g/mol

Mass of silver chloride formed = 8.4372484 g

Mass of silver chloride formed ≅ 8.44 g

Hence, the mass of silver chloride (AgCl) formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride is 8.44 g. The correct option is the second option 8.44 g

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Answer : The mass of nitrogen present are, 4.965 grams.

Explanation :

According to the Raoult's law,

[tex]p_{He}=X_{He}\times p_T[/tex]

where,

[tex]p_{He}[/tex] = partial pressure of gas = 231 mmHg

[tex]p_T[/tex] = total pressure of gas = 641 mmHg

[tex]X_{He}[/tex] = mole fraction of helium gas = ?

Now put all the given values in this formula, we get the mole fraction of helium gas.

[tex]231mmHg=X_{He}\times 641mmHg[/tex]

[tex]X_{He}=0.36[/tex]

Now we have to calculate the mole fraction of nitrogen gas.

[tex]X_{He}+X_{N_2}=1[/tex]

[tex]X_{N_2}=1=0.36=0.64[/tex]

Now we have to calculate the mass nitrogen gas.

[tex]\frac{X_{He}}{X_{N_2}}=\frac{n_{He}}{n_{N_2}}[/tex]

[tex]\frac{X_{He}}{X_{N_2}}=\frac{\frac{w_{He}}{M_{He}}}{\frac{w_{N_2}}{M_{N_2}}}[/tex]

where,

n = moles, w = mass, M = molar mass

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

[tex]\frac{0.36}{0.64}=\frac{\frac{0.399}{4}}{\frac{w_{N_2}}{28}}[/tex]

[tex]w_{N_2}=4.965g[/tex]

Therefore, the mass of nitrogen present are, 4.965 grams.

Answer:

0.25 mol of water

Explanation:

The starting point is writing out a balanced chemical equation

[tex]2LiHCO_{3} --->Li_{2}O +2CO_{2} +H_{2}O[/tex]

From here we can see that 2 moles of Lithium hydrogen carbonate produce 1 mole of water. It follows that 0.5 moles of Lithium hydrogen carbonate will produce half of that too, which is 0.25 moles. The ratio is always mantained.

Final answer:

When 0.50 mol of lithium hydrogen carbonate (LiHCO3) decomposes, it produces 0.25 mol of water (H2O).

Explanation:

You asked how many moles of H2O are formed when 0.50 mol LiHCO3 decomposes. The decomposition reaction for LiHCO3 to produce Li2O, CO2, and H2O is:

2 LiHCO3(s) → Li2O(s) + 2 CO2(g) + H2O(g)

According to the balanced equation, 2 moles of LiHCO3 produce 1 mole of H2O. Therefore, when 0.50 mole of LiHCO3 decomposes, you would get half of that ratio in moles of water:

(0.50 mol LiHCO3) / (2 mol LiHCO3) = 0.25 mol H2O

Therefore, the decomposition of 0.50 mol of LiHCO3 will produce 0.25 mol of H2O.

Answer:

For 1: The volume of carbon monoxide required to produce given amount of methanol is 11.9 L.

For 2: The volume of oxygen required to form given amount of water is 7.77 L.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     ......(1)

For methanol:

Given mass of methanol = 19.3 g

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of methanol}=\frac{19.3g}{32g/mol}=0.603mol[/tex]

The chemical reaction of formation of methanol from carbon monoxide follows:

[tex]CO+2H_2\rightarrow CH_3OH[/tex]

By Stoichiometry of the reaction:

1 mole of methanol is produced from 1 mole of carbon monoxide.

so, 0.603 moles of methanol will be produced from = [tex]\frac{1}{1}\times 0.603=0.603mol[/tex] of carbon monoxide.

To calculate the volume of carbon monoxide, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of carbon monoxide = 676 mmHg

V = Volume of carbon monoxide = ? L

n = Number of moles of carbon monoxide = 0.603 mol

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

T = Temperature of carbon monoxide = 212 K

Putting values in above equation, we get:

[tex]676mmHg\times V=\frac{19.3}{32g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 212K\\\\V=11.79L[/tex]

Hence, the volume of carbon monoxide required to produce given amount of methanol is 11.9 L.

Calculating the moles of water by using equation 1, we get:

Given mass of water = 12.5 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{12.5g}{18g/mol}=0.694mol[/tex]

The chemical reaction of formation of water from oxygen and hydrogen follows:

[tex]O_2+2H_2\rightarrow 2H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of water is produced from 1 mole of oxygen gas.

so, 0.694 moles of water will be produced from = [tex]\frac{1}{2}\times 0.694=0.347mol[/tex] of oxygen gas.

At STP:

1 mole of a gas occupies 22.4 L of volume.

So, 0.347 moles of oxygen gas will occupy = [tex]\frac{22.4L}{1mol}\times 0.347mol=7.77L[/tex]

Hence, the volume of oxygen required to form given amount of water is 7.77 L.

Answer: Aluminum, [tex]2.70 g/cm^3[/tex]

Explanation:

Density is defined as the mass contained per unit volume.  It is characteristic of a substance.

[tex]Density=\frac{mass}{Volume}[/tex]

Given : Mass of object = 8.44 grams

Volume of object=[tex]length\times breadth\times height=1.25cm\times 2.50cm\times 1.0cm=3.125cm^3[/tex]

Putting in the values we get:

[tex]Density=\frac{8.44g}{3.125cm^3}=2.70g/cm^3[/tex]

Thus density of the object will be [tex]2.70 g/cm^3[/tex]  which matches that of aluminium.

Answer: Nickel

Explanation:

They have the same density as each other

Answer:

2. mass number

Explanation:

Mass number is equal to the sum of protons + neutrons in an element. For silicon is 14 + 14 = 28 (for the silicon 14 isotope).

Answer:

2. mass number

Explanation:

The atomic mass is concentrated in the tiny nucleus which contains protons and neutrons. These particles are called the nucleons. Nucleons dictate the mass of an atom.

The number of protons and neutrons in a nucleus is the atomic mass. Also, when we add the protons and neutrons together, we are calculating the mass number. The mass number also gives the atomic mass.

Atomic number on the other hand is the number of electrons or protons in a neutral atom.

Answer :  The partial pressure of oxygen is, 799.19 torr

Solution :

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

[tex]P_T=p_{H_2O}+p_{O_2}[/tex]

where,

[tex]P_T[/tex] = total partial pressure = 805 torr

[tex]P_{O_2}[/tex] = partial pressure of oxygen gas = ?

[tex]P_{H_2O}[/tex] = partial pressure of water = 25.81 mm Hg = 25.81 torr

Now put all the given values is expression, we get the partial pressure of the oxygen gas.

[tex]805\text{ torr}=25.81\text{ torr}+p_{O_2}[/tex]

[tex]p_{O_2}=779.19\text{ torr}[/tex]

Therefore, the partial pressure of oxygen gas is, 799.19 torr

Answer: Option (b) and (d) are correct.

Explanation:

An equilibrium reaction is defined as the reaction in which rate of forward reaction equals rate of backward reaction.

A photosynthesis reaction is the reaction in which plants in the presence of sunlight, water, and carbon dioxide make their own food.  

Hence, the statement rate of formation of [tex]O_{2}[/tex] is equal to the rate of formation of [tex]CO_{2}[/tex] is true.

        Hence, the statement concentrations of [tex]CO_{2}[/tex] and [tex]O_{2}[/tex] will not change, is true.

Some amount of carbon dioxide might escape out into the air.

Hence, the statement concentrations of [tex]CO_{2}[/tex] and [tex]O_{2}[/tex] will be equal, is false.

Hey there!:

To get to the molar solubility from the Ksp, you have to think about the how Ksp is calculated, and what it means.  

For a compound that forms 2 ions, Ksp = X^2 where X is the molar solubility. For a compound that forms 3 ions, Ksp = 4X^3, where again, X is the molar solubility.  

If you calculate the molar solubilities of each of your compounds, you will see that  A2X  has the higher molar solubilty

Hope this helps!

Answer:

A₃X

Explanation:

In order to find the molar solubility (S) of a compound, we will use an ICE Chart.

A₂X

Let's consider the solution of A₂X.

     A₂X(s) ⇄ 2 A⁺(aq) + X²⁻(aq)

I                         0           0

C                     +2S        +S

E                       2S          S

The solubility product (Kps) is:

Kps = 1.5 × 10⁻⁵ = [A⁺]².[X²⁻] = (2S)².S = 4S³

S = 0.016 M

A₃X

Let's consider the solution of A₃X.

     A₃X ⇄ 3 A⁺(aq) + X³⁻(aq)

I                    0             0

C                 +3S         +S

E                   3S           S

The solubility product (Kps) is:

Kps = 1.5 × 10⁻⁵ = [A⁺]³.[X³⁻] = (3S)³.S = 27 S⁴

S = 0.027 M

A₃X has a higher molar solubility than A₂X.

Answer:

[tex]4.4285\times 10^8 [/tex] many atoms would it take to make the distance 6.20 cm from end to end.

Explanation:

Diameter of the helium atom =  [tex]d=1.40\times 10^2 pm[/tex]

Let the number of atoms required to make the distance 6.20 cm be n.

[tex]d\times n= 6.20 cm =6.20\times 10^{10} pm[/tex]

[tex]1.40\times 10^2 pm\times n=6.20\times 10^{10} pm[/tex]

[tex]n=\frac{6.20\times 10^{10} pm}{1.40\times 10^2 }=4.4285\times 10^8 [/tex]

[tex]4.4285\times 10^8 [/tex] many atoms would it take to make the distance 6.20 cm from end to end.

To find the number of helium atoms needed to span 6.20 cm, convert the atom's diameter to centimeters and divide 6.20 cm by the atom's diameter. After calculation, you will get the number of helium atoms required.

We first need to convert the diameter of a helium atom from picometers to centimeters: 1.40 × 102 pm is equivalent to 1.40 × 10-10 cm. Then, we divide the total length required, 6.20 cm, by the diameter of one helium atom in centimeters to find the number of atoms required:

Number of atoms = Total distance / Diameter of one atom = 6.20 cm / (1.40 × 10-10 cm)

After calculating this expression, we would arrive at the number of helium atoms needed to span the distance of 6.20 cm.