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vovangra
9 days ago
10

Imagine you are volunteering in a 1st grade classroom. The students recently learned about the water cycle. One student raises h

er hand and asks, "I heard that elements like carbon can cycle too. How does carbon cycle? Is it like how water cycles?" How would you explain the cycling of carbon to this 1st grade class? Keep in mind that the students have likely not taken an advanced science course yet!
Chemistry
1 answer:
Alekssandra [2.8K]9 days ago
5 0
The carbon cycle plays a crucial role in sustaining life on Earth. In this cycle, carbon dioxide from the atmosphere is absorbed by plants through their stomata for photosynthesis. When animals consume this plant matter, carbon transfers into their bodies. Animals release carbon dioxide back into the atmosphere during respiration.
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A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state.Th
castortr0y [2916]
The answer is C. The specific amount of energy released when excited electrons fall back to the ground state produces an emission spectrum. That energy is emitted as photons with precise wavelengths corresponding to the energy differences between levels. Because each element yields a characteristic set of wavelengths, the emission spectrum can be used to identify the element in the sample.
5 0
16 days ago
Read 2 more answers
Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
eduard [2645]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K

Starting temperature of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

mC \Delta T = mC \times \Delta T

10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

T_{f} - 100^{o}C = 0^{o}C - T_{f}

2T_{f} = 100^{o}C

T_{f} = 50^{o}C

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

\Delta H = mC \Delta T

= 10000 g \times 0.385 J/K g \times 323 K

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

= 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

= 10000 g \times 0.385 J/K g \times -0.143

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

                   = -554.12 J/K + 647.49 J/K\Delta S_{total} = \Delta S_{1} + \Delta S_{2}

           = 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

6 0
14 days ago
A certain liquid has a density of 2.67 g/ cm3. what is the mass of 30.5 ml of this liquid? (
lorasvet [2668]
Hello!

density = 2.67 g/cm³

volume = 30.5 mL

Thus:

Mass = density * volume

Mass = 2.67 * 30.5

Mass = 81.435 g 
4 0
1 month ago
The table lists the lattice energies of some compounds. Compound Lattice Energy (kJ/mol) LiF –1,036 LiCl –853 NaF –923 KF –821 N
KiRa [2857]

Response: Option A) The lattice energy rises as cations become smaller, as demonstrated by LiF and KF.

Clarification: It has been observed that the lattice energy is largely determined by two primary factors regarding ionic solids:

i) The ionic charges - An increase in the charge of the ions corresponds to an increase in lattice energy.

and

ii) The size or radius of the ions - As the ionic size grows, the lattice energy diminishes accordingly.

Therefore, in this context, the latter factor is evident. Thus, it can be concluded that as cation sizes decrease among ionic solids, the lattice energy increases.

7 0
12 days ago
Read 2 more answers
If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point o
alisha [2865]
<span>Some solutions demonstrate colligative properties, which rely on the quantity of solute in a solvent. To find the elevation in boiling point, we use the formula:

</span><span>ΔT(boiling point)  = (Kb)mi

where Kb represents a constant, m is the solution's molality, and i is the van't Hoff factor.

From the provided information, we can easily determine i as follows:

</span>ΔT(boiling point)  = (Kb)mi
103.45 - 100  = (0.512)3.90i
i = 1.73 <-------van't Hoff factor
7 0
1 month ago
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