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3 years ago

In chemistry class what do we mean when we refer to a mole

1 answer:

mote1985 [20]3 years ago

6 0

A mole is just a unit of measurement. It is basically a <span>quantity of anything that has the same number of particles found in 12.000 grams of carbon-12. </span>

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Electron A falls from energy level X to energy level Y and releases blue light. Electron B falls from energy level Y to energy l

MrMuchimi

Answer: Transition from X to Y will have greater energy difference.

Explanation: For studying the energy difference, we require Planck's equation.

$E=\frac{hc}{\lambda}$

where, h = Planck's Constant

c = Speed of light

E = Energy

$\lambda$ = Wavelength of particle

From the equation, it is visible that the energy and wavelength follow inverse relation which means that with low wavelength value, energy will be the highest and vice-versa.

As electron A falls from X-energy level to Y-energy level, it releases blue light which has low wavelength value (around 470 nm) which means that it has high energy.

Similarly, Electron B releases red light when it falls from Y-energy level to Z-energy level, which has high wavelength value (around 700 nm), giving it a low energy value.

Energy Difference between X-energy level and Y-energy level will be more.

5 0

3 years ago

Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.

Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0

3 years ago

Read 2 more answers

Which compound has the highest percent composition by mass of strontium?

spayn [35]

The answer is (3). The number of Sr is the same so if the compound has the smallest gram formula mass, it has the highest percent composition by mass of strontium. So the answer is (3).

5 0

4 years ago

Read 2 more answers

A container of 56mL is at STP, The pressure is increased to 3.6atm and the volume to 162 mL.

Sindrei [870]

Answer:

T2 = 2843.1 oK. This is a huge temperature. Check it for errors.

Explanation:

Remark

This is the same question as the other one I've answered. Only the numbers have been altered.

Givens

v1 = 56 mL

P1 = 1 atm

T1 = 273o K

v2 = 162

P2 = 3.6 atm

T2 = ?

Formula

Vi * P1 / T1 = V2 * P2/T2

Solution

Rearrange the formula so T2 is on the left

T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.

T2 = 162 * 3.6* 273 / (56 *1)

T2 = 159213.6/56

T2 = 2843.1

3 0

3 years ago

Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. neutraliz

madreJ [45]

Answer:

0.197 M

Explanation:

The reaction equation is:

H2SO4(aq) +2KOH(aq) ----> K2SO4(aq) + 2H2O(l)

number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles

number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles

since H2SO4 is the reactant in excess;

2 moles of KOH reacts with 1 mole of H2SO4

0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4

Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles

Total volume = 0.25 L+ 0.2 L = 0.45 L

concentration of H2SO4 = 0.0885/0.45 = 0.197 M

4 0

3 years ago