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3 years ago

Does the suns light exhibit an emission spectrum?

1 answer:

algol133 years ago

7 0

Light is emitted in the visible spectrum to nearly all energies, so in the spectrum of the Sun you can see all colours. This is clearly not a continuous spectrum, but still represents visible radiation, divided into its component colours.

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How many moles of iron is 6.022 x 10^22 atoms of iron? (Report answer as a number rounded to one place past the decimal.) *

yuradex [85]

<h3>Answer:</h3>

$\displaystyle 0.1 \ mol \ Fe$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

6.022 × 10²² atoms Fe (iron)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 6.022 \cdot 10^{22} \ atoms \ Fe(\frac{1 \ mol \ Fe}{6.022 \cdot 10^{23} \ atoms \ Fe})$
Divide: $\displaystyle 0.1 \ mol \ Fe$

7 0

3 years ago

What is the percentage by volume of 30.0 mL ethanol (C2H6O) dissolved in 150.0 mL water (H2O)?

NeX [460]

Answer: 16.7 %

Explanation:

$\frac{30.0}{150.0+30.0}=\boxed{16.7\%}$

5 0

2 years ago

Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

3 years ago

Convert 364 kg to lb

miv72 [106K]

Answer:

This snip might help...it depends :)

Explanation:

5 0

3 years ago

Transition metals can be found in Families 3 through 12 on the periodic table. What property do transition metals share?

Vedmedyk [2.9K]

The answer might be C

8 0

3 years ago

Read 2 more answers