All categories

3 years ago

During an endothermic phase change, what happens to the potential energy and the kinetic energy?

2 answers:

7 0

The correct answer should be that Potential energy increases, and kinetic energy increases, since they both increase as the temperature changes.

m_a_m_a [10]3 years ago

7 0

Answer:

D.) Potential energy increases, and kinetic energy stays the same

Explanation:

edge 2020

hope this helps!!

You might be interested in

What atomic or hybrid orbitals make up the sigma bond between Cl and F in chlorine trifluoride, ClF3

Umnica [9.8K]

Answer:

sp³d¹ hybridization

Explanation:

Given Cl as central element with three F substrates ...

The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.

Valence bond theory predicts the following during bonding:

Cl:[Ne]3s²3p²p²p¹3d⁰

=> [Ne]3s²p²p¹p¹d¹

=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹

giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.

Note the following images:

Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.

7 0

3 years ago

an element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with

ryzh [129]

Answer is: the average atomic mass 217.606 amu.

Ar₁= 203.973 amu; the average atomic mass of isotope.

Ar₂ = 205.9745 amu.

Ar₃ = 206.9745 amu.

Ar₄ = 207.9766 amu.

ω₁ = 1.40% = 0.014; mass percentage of isotope.

ω₂ = 24.10% = 0.241.

ω₃ = 22.10% = 0.221.

ω₄ = 57.40% = 0.574.

Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.

Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

Ar = 217.606 amu.

But abundance of isotopes is greater than 100%.

It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.

7 0

3 years ago

When pellets of NaOH(s) are added to a flask of water, it is observed that the temperature of the water increases as the pellets

Nataly [62]

Answer: It is an exothermic process because heat energy is absorbed by the water as the NaOH(s) dissolves in it.

Explanation:

Endothermic reaction : It is a type of chemical reaction where the energy is absorbed from the surrounding. In the endothermic reaction, the reactant are less than the energy of product. In endothermic reaction, the change in enthalpy is, positive

Exothermic reaction : It is a type of chemical reaction where the energy is released into the surrounding. In the exothermic reaction, the energy of reactant are more than the energy of product. In exothermic reaction, the change in enthalpy is, negative.

Thus as energy is absorbed by water, which has been released by NaOH, the process is exothermic.

3 0

3 years ago

Melatonin has the chemical formula C13H16N2O2. Less than 10 milligrams(0.01g) Convert the melatonin to moles

oee [108]

Answer: There are 0.000043 moles of melatonin.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

given mass = 10 mg = 0.01 g

molar mass of $C_{13}H_{16}N_2O_2$ = 232.28 g

${\text {moles of melatonin}}=\frac{0.01g}{232.28g/mol}=0.000043moles$

Thus there are 0.000043 moles of melatonin

6 0

3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago