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IgorLugansk [536]

3 years ago

9

When looking up at the clouds, you can usually make out different shapes and figures with _____. cumulus clouds stratus clouds c

irrus clouds altostratus clouds
Desert climates can have very hot days and very cold nights. This is because _____. there is usually cloud coverage there is usually a clear sky there is usually precipitation there is usually hot or cold wind

1 answer:

MAXImum [283]3 years ago

4 0

The answer to the question stated above is cumulus clouds.
<span>When looking up at the clouds, you can usually make out different shapes and figures with cumulus clouds.
</span>Cumulus <span>low-level clouds. They </span>are the puffy, white, cotton-top clouds that look so soft.

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How can one kg of iron melt more ice than 1 kg lead at 100 °C

Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{iron}$ = Heat obtained from the iron by the ice

ΔQ $_{iron}$ = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{lead}$ = Heat obtained from the iron by the ice

ΔQ $_{lead}$ = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0

4 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

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4 years ago

A rock is an <br> of minerals.

Natasha_Volkova [10]

Rocks are minerals, made up of many minerals. There are 3 types of rocks which are Igneous, Sedementary, and Metamorphic.

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3 years ago

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The smallest ionic radius among the following. <br> Li<br> Mg<br> Na+<br> be+

dimaraw [331]

The correct answer is Be+

That is because it lost a single electron but still has the same number of protons, and thus the effective charge attracting each electron is greater, which in turn makes the radius even smaller

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3 years ago