State one function of Emitter??

Explain how you would prepare a solution of sucrose with a molarity of 0.25.

Dimas [21]

Solution of 0.25 M is prepared in two steps,

1) Calculate Amount of Solute:

Molar Mass of Solute: 342.3 g/mol

As we know,
Molarity = Moles / 1 dm³
or,
Moles = Molarity × 1 dm³
Putting Values,
Moles = 0.25 mol.dm⁻³ × 1 dm³

Moles = 0.25 moles

Now, find out mass of sucrose,
As,
Moles = Mass / M.mass
or,
Mass = Moles × M.mass
Putting Values,
Mass = 0.25 mol × 342.3 g.mol⁻¹

Mass = 85.57 g

2) Prepare Solution:
Take Volumetric flask and add 85.57 g of sucrose in it. Then add distilled water up to the mark of 1 dm³. Shake well! The solution prepared is 0.25 M in 1 Liter.

4 0

3 years ago

Deep sea divers use a mixture of helium and oxygen to breathe. Assume that a diver is going to a depth of 150 feet where the tot

Svetlanka [38]

Answer:

4.525% is the percentage by volume of oxygen in the gas mixture.

Explanation:

Total pressure of the mixture = p = 4.42 atm

Partial pressure of the oxygen = $p_1=0.20 atm$

Partial pressure of the helium = $p_2$

$p_1=p\times \chi_1$ (Dalton law of partial pressure)

$0.20 atm=4.42 atm\times \chi_1$

$\chi_1=\frac{0.20 atm}{4.42 atm}=0.04525$

$\chi_2=1-\chi_1=1-0.04525=0.95475$

$chi_1+chi_2=1$

$n_1=0.04525 mol,n_2=0.95475 mol$

According Avogadro law:

$Moles\propto Volume$ (At temperature and pressure)

Volume occupied by oxygen gas = $V_1$

Total moles of gases = n = 1 mol

Total Volume of the gases = V

$\frac{n_1}{V_1}=\frac{n}{V}$

$\frac{V_1}{V}=\frac{n_1}{n}=\frac{0.04525 mol}{1 mol}$

Percent by volume of oxygen in the gas mixture:

$\frac{V_1}{V}\times 100=\frac{0.04525 mol}{1 mol}\times 100=4.525\%$

6 0

3 years ago

What else supports the idea that an exothermic reaction has occurred?

kodGreya [7K]

The main pieces of evidence that an exothermic reaction has occurred is an increase in temperature due to the release of energy, a release of energy in the form of light, or a release of gas.

4 0

3 years ago

Consider the reaction: I2(g)+Cl2(g)⇌2ICl(g) Kp= 81.9 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each condition: -

scoray [572]

Answer:

Part 1: - 1.091 x 10⁴ J/mol.

Part 2: - 1.137 x 10⁴ J/mol.

Explanation:

Part 1: At standard conditions:

At standard conditions Kp= 81.9.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.

Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.

For the reaction:

I₂(g) + Cl₂(g) ⇌ 2ICl(g).

Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.

6 0

3 years ago

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The d-metals iron, copper, and manganese form cations with different oxidation states. For this reason, they are found in many o

choli [55]

Answer:

They have electrons in their 3d- and 4s-orbital for bond formation.

Explanation:

d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.

The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.

If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.

If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2

4 0

3 years ago

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State one function of Emitter??​

State one function of Emitter??