To make solid barium sulfide, you would need to react barium metal with elemental sulfur. The balanced chemical equation for this reaction is:
Ba(s) + S(s) → BaS(s)
To carry out this reaction, you would need to add excess sulfur to the barium metal. This ensures that all the barium is consumed in the reaction, and no excess barium remains. The excess sulfur can be removed by washing the product with a suitable solvent.
It is important to note that the reaction between barium and sulfur can be exothermic, releasing heat and potentially causing a fire or explosion. Therefore, appropriate safety precautions, such as wearing gloves and eye protection and working in a well-ventilated area, should be taken when carrying out this reaction.
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To make a solid barium sulfide (BaS) you would need to add sulfur (S) to barium (Ba) in a stoichiometric ratio of 1:1. This means that for every one mole of barium, you would need one mole of sulfur.
The reaction can be represented by the following chemical equation:
Ba + S → BaS
To carry out this reaction, you could start with a sample of metallic barium and add elemental sulfur powder to it, in a ratio of 1:1 by mole. The reaction between the two elements will produce solid barium sulfide.
It is important to note that this reaction can be highly exothermic, so appropriate safety precautions should be taken. Additionally, barium sulfide is a toxic and reactive compound, and should be handled with care.
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a sample of nobr was placed on a 1.00l flask containing no no or br 2 at equilibrium the flask contained
At equilibrium, the concentrations of NO, Br2, and NOBr in the flask will remain constant. However, without specific values for the initial concentration of NOBr or the equilibrium constant (Kc), it's not possible to determine.
.Based on the provided information, it seems that a sample of NOBr was placed in a 1.00 L flask at equilibrium, which means that the NOBr has decomposed into NO and Br2.
At equilibrium, the concentrations of NO, Br2, and NOBr in the flask will remain constant. However, without specific values for the initial concentration of NOBr or the equilibrium constant (Kc), it's not possible to determine the exact concentrations of these substances in the flask.
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A sample of NOBr being placed in a 1.00 L flask containing no NO or Br2 at equilibrium, I'll first provide the balanced chemical equation for the reaction:
[tex]2 NOBr (g) ⇌ 2 NO (g) + Br2 (g)[/tex]
At equilibrium, the concentrations of the reactants and products remain constant. To determine the concentrations of NOBr, NO, and Br2 at equilibrium, we need to follow these steps:
1. Write the expression for the equilibrium constant (Kc) based on the balanced chemical equation:
[tex]Kc = [NO]^2 [Br2] / [NOBr]^2[/tex]
2. Set up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of the species involved in the reaction. The initial concentrations of NO and Br2 are 0 since they are not initially present in the flask.
NOBr NO Br2
I C0 0 0
C -2x +2x +x
E C0-2x 2x x
3. Substitute the equilibrium concentrations from the ICE table into the Kc expression:
[tex]Kc = (2x)^2 * x / (C0-2x)^2[/tex]
4. To solve for x, you need the value of Kc for the reaction. Look up the Kc value for this reaction in a reference or use provided information. Once you have Kc, substitute it into the equation and solve for x.
5. Calculate the equilibrium concentrations of NOBr, NO, and Br2 by substituting the value of x back into the ICE table:
[NOBr] = C0-2x
[NO] = 2x
[Br2] = x
By following these steps, you can determine the concentrations of NOBr, NO, and Br2 in the 1.00 L flask at equilibrium.
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PLEASE ANSWER ASAP
1. How many atoms are present in 8.500 mole of chlorine atoms?
2. Determine the mass (g) of 15.50 mole of oxygen.
3. Determine the number of moles of helium in 1.953 x 108 g of helium.
4. Calculate the number of atoms in 147.82 g of sulfur.
5. Determine the molar mass of Co.
6. Determine the formula mass of Ca3(PO4)2.
IT WOULD BE HELPFUL
1) 5.1167 x 10²⁴atoms of chlorine. 2) 248.00 g. 3) 4.8825 x 10⁷ moles of helium. 4) 2.7757 x 10²⁴ atoms of sulfur. 5) Molar mass of Co (cobalt) is 58.93 g/mol. 6) Formula mass = 310.18 g/mol.
What is meant by formula mass?Sum of the atomic masses of all the atoms in chemical formula is called formula mass
1.) Number of atoms = 8.500 moles x 6.022 x 10²³ atoms/mole = 5.1167 x 10²⁴ atoms of chlorine.
2.) Molar mass of oxygen is 16.00 g/mol. Therefore:
Mass of 15.50 moles of oxygen = 15.50 moles x 16.00 g/mol = 248.00 g.
3.) Molar mass of helium is 4.00 g/mol. Therefore, the number of moles of helium in 1.953 x 10⁸ g is:
Number of moles = 1.953 x 10⁸ g / 4.00 g/mol = 4.8825 x 10⁷ moles of helium.
4.) Molar mass of sulfur is 32.06 g/mol. Therefore, the number of moles of sulfur in 147.82 g is:
Number of moles = 147.82 g / 32.06 g/mol = 4.6084 moles of sulfur.
To find the number of atoms, we can use Avogadro's number again:
Number of atoms = 4.6084 moles x 6.022 x 10²³ atoms/mole = 2.7757 x 10²⁴ atoms of sulfur.
5.) Molar mass of Co (cobalt) is 58.93 g/mol.
6.) Ca₃(PO₄)₂ contains 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms.
Atomic masses of these elements are:
Calcium (Ca) = 40.08 g/mol
Phosphorus (P) = 30.97 g/mol
Oxygen (O) = 16.00 g/mol
Therefore, formula mass of Ca₃(PO₄)₂ is:
Formula mass = (3 x 40.08 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)
= 120.24 g/mol + 61.94 g/mol + 128.00 g/mol
= 310.18 g/mol.
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The cloud droplets in a cloud are formed by water vapor molecules and: A) protons. B) ions. C) molecules of air. D) condensation nuclei.
Answer:
condensation nuclei
Explanation:
What is the concentration (in molality) of an aqueous solution of NaCl made by adding
4.56 g of NaCl to enough water to give 20.0 mL of solution. Assume the density of the
solution is 1.03 g/mL
Answer:
data given
mass of NaCl 4.56
dissolved volume 20ml(0.02l)
density of solution 1.03g/ml
Required molality
Explanation:
molarity=m/mr×v
where
m is mass
mr molar mass
v is volume
now,
molarity=4.56/58.5×0.02
molarity =3.9
: .molarity is 3.9mol/dm^3
According to molal concentration, the concentration (in molality) of an aqueous solution of NaCl is 0.0047 mole/kg.
What is molal concentration?Molal concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molal concentration is moles/kg.
The molal concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molal concentration is calculated by the formula, molal concentration=mass/ molar mass ×1/mass of solvent in kg.
In terms of moles, it's formula is given as molal concentration= number of moles /mass of solvent in kg.
Substitution in formula gives the answer but first mass of solution is determined which is density×volume= 1.03×20=20.6 g , mass of solvent= 20.6-4.56=16.05, thus molal concentration=4.56/58.5×1/16.05=0.0047 moles/kg.
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how many moles of naf must be dissolved in 1.00 liter of a saturated solution of pbf2 at 25˚c to reduce the [pb2 ] to 1 x 10–6 molar? (ksp pbf2 at 25˚c = 4.0 x 10–8)
The moles of NaF that must be dissolved in 1.00 liter of a saturated solution of PbF₂ at 25˚C to reduce the [Pb²⁺] to 1 x 10⁻⁶ molar is 2.0 x 10⁻⁵.
The solubility product expression for PbF₂ is given by:
Ksp = [Pb²⁻][F-]²At equilibrium, the product of the ion concentrations must be equal to the solubility product constant. We are given that the [Pb²⁺] in the saturated solution is 1 x 10⁻⁶ M. Therefore, we can use the Ksp expression to calculate the concentration of F- in the solution:
Ksp = [Pb²⁺][F⁻]²4.0 x 10⁻⁸ = (1 x 10⁻⁶)([F⁻]²)[F⁻]² = 4.0 x 10⁻²[F⁻] = 2.0 x 10⁻¹Now, we can calculate the amount of NaF needed to reduce the [F⁻] concentration to 2.0 x 10⁻¹ M. Since NaF is a 1:1 electrolyte, the concentration of F- will be equal to the concentration of NaF added.
Number of moles of NaF = (2.0 x 10⁻¹) mol/L x 1.00 L = 2.0 x 10⁻¹ molesHowever, we need to dissolve this amount of NaF in a saturated solution of PbF₂. Therefore, we need to check that the amount of NaF we added will not exceed the maximum amount that can dissolve in the solution at 25˚C.
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you have 400 grams (g) of a substance with a half life of 10 years. how much is left after 100 years?
After 100 years, there will be 6.25 grams of the substance remaining.
What is half life?Half-life is the time it takes for half of the radioactive atoms in a sample to decay or for the concentration of a substance to decrease by half.
Amount remaining = initial amount x (1/2)^(number of half-lives)
In this case, half-life of the substance is 10 years, which means that after 10 years, half of the substance will have decayed. After another 10 years (20 years total), half of remaining substance will decay, leaving 1/4 of the original amount. After another 10 years (30 years total), half of that remaining amount will decay, leaving 1/8 of the original amount. This process continues every 10 years.
To find the amount of substance remaining after 100 years, we need to know how many half-lives have occurred in that time: 100 years / 10 years per half-life = 10 half-lives
Amount remaining = 400 g x (1/2)¹⁰= 6.25 g
Therefore, after 100 years, there will be 6.25 grams of the substance remaining.
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a 1.25 g sample of co2 is contained in a 750. ml flask at 22.5 c. what is the pressure of the gas, in atm?
The pressure of gas is 1.05 atm when a 1.25 g sample of CO₂ is contained in a 750ml flask at 22.5°C.
Molecular weight of CO₂ is 1.25g ,Volume of CO₂ is 750ml,Temperature of CO₂ is 22.5°C and the gas constant is 0.08206 L atm/mol K.
Using the ideal gas law equation the pressure is found to be 1.05 atm.
To calculate the pressure of the gas, we can use the ideal gas law equation: [tex]PV=nRT[/tex]
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the volume to liters by dividing by 1000: 750 ml = 0.75 L.
Next, we need to calculate the number of moles of CO₂ present in the flask. We can use the molecular weight of CO₂ to convert from grams to moles:
[tex]1.25 * (1 /44.01 ) = 0.0284 mol[/tex]
Now we can plug in the values into the ideal gas law equation:
[tex]PV=nRT[/tex]
[tex]P * 0.75 L = 0.0284 mol * 0.08206 L*atm/mol*K * (22.5 + 273.15) K[/tex]
Simplifying and solving for P, we get:
[tex]P = (0.0284 * 0.08206 * 295.65) / 0.75 = 1.05 atm[/tex]
Therefore, the pressure of the gas in the flask is 1.05 atm.
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