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dimulka [17.4K]

1 year ago

15

Write a hypothesis for Section 2 of the lab, which is about the effect of the angle of insolation on the amount of heat a materi

al absorbs. Use the format "if . . . then . . . because . . ." and be sure to answer this part of the lesson question: "What factors influence the absorption of sunlight at Earth's surface?"

1 answer:

m_a_m_a [10]1 year ago

5 0

Answer:

If the angle of insolation is higher, then the temperature of the soil will be higher because it will receive more direct light.

Explanation:

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What is the enthalpy for reaction 1 reversed?reaction 1 reversed: N2O4→N2 + 2O2

Nutka1998 [239]

Answer:

8kJ/mol

Explanation:

since the forward reaction is -8kJ/mol, the backward reaction has the same enthalphy but reversed

7 0

1 year ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

1 year ago

Identify the term that applies to each definition.

Illusion [34]

Answer:

A. Reference blank

B. Cuvettes

C. Transmittance

D. Absorbance

E. Wavelength

Explanation:

A reference blank is a sample prepared using the solvent and any other chemicals in the sample solutions, but not the absorbing substance.

A square-shaped container, typically made of quartz, designed to hold samples in a spectrophotometer is known as Cuvettes.

A measurement of the amount of light that passes through a sample or percentage of light transmitted by the sample, with the respective intensities of the incident and transmitted beams is called Transmittance.

The measurement of the amount of light taken in by a sample is known as Absorbance

The wavelength is also the distance travelled by the wave during a period of oscillation. In spectrophotometry, the unit is inversely proportional to energy and commonly measured in nanometers

6 0

1 year ago

how many moles of cesium carbonate will be produced when 5.34 moles of cesium reacts with iron iii carbonate

irga5000 [103]

Answer: 2.7moles

Explanation:

6Cs + Fe2(CO3)3 —> 3Cs2CO3 + 2Fe

From the equation,

6moles of Cs produced 3 moles of Cs2CO3.

Therefore, 5.34 moles of Cs will produce = (5.4x3)/6 = 2.7moles of Cs2CO3

6 0

2 years ago

Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t

-Dominant- [34]

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒ ABCC = $\frac{4RBCC}{\sqrt{3} }$

⇒ = $\frac{4\times 0.12584}{\sqrt{3}}$

⇒ = $0.29062 \ nm$

Likewise AFCC as well as RFCC are interconnected by

√2AFCC = 4RFCC

⇒ AFCC = $\frac{4RFCC}{\sqrt{2}}$

⇒ = $\frac{4\times 0.12894}{\sqrt{2} }$

⇒ = $0.36470 \ nm$

Now,

The Change in Percent Volume,

= $\frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent$

= $\frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent$

= $\frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent$

= $\frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent$

= $97.62 \ percent (approximately)$

Note: percent = %

7 0

1 year ago