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ohaa [14]
3 years ago
13

When light shines on a sample, each element emits specific wavelengths producing a unique fingerprint called its ______ spectra.

A) infrared B) line C) raman D) ultraviolet
Chemistry
2 answers:
nata0808 [166]3 years ago
7 0

Answer:

B) line is the correct answer.

Explanation:

When light shines on a sample, each element emits specific wavelengths producing a unique fingerprint called its line spectra.

The line spectrum is electromagnetic spectra consist of discrete spectra lines.

When the atoms are excited they emit ray of specific wavelengths that correspond to various colors. The emitted ray can be seen as a range of colored line, this range of colored lines is termed as line spectra.

lines spectra are usually utilized to recognize atoms and molecules. Each element has individual line emission spectra.

DanielleElmas [232]3 years ago
3 0

Answer:

B.

Explanation:

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The volume of a sample of oxygen is 200.0 mL when the pressure is 3.000 atm and the temperature is 37.0 oC. What is the new temp
attashe74 [19]

Answer:

Explanation:

Given:

V1 = 200 ml

T1 = 20 °C

= 20 + 273

= 293 K

P1 = 3 atm

P2 = 2 atm

V2 = 400 ml

Using ideal gas equation,

P1 × V1/T1 = P2 × V2/T2

T2 = (2 × 400 × 293)/200 × 3

= 234400/600

= 390.67 K

= 390.67 - 273

= 117.67 °C

7 0
3 years ago
Determine the rate of a reaction that follows the rate law rate = k a m b i where
ikadub [295]
(missing part of your question):
when we have K = 1 x 10^-2 and [A] = 2 M & [B] = 3M & m= 2 & i = 1
So when the rate = K[A]^m [B]^i
and when we have m + i = 3 so the order of this reaction is 3 So the unit of K is L^2.mol^-2S^-1
So by substitution:
∴ the rate = (1x 10 ^-2 L^-2.mol^-2S^-1)*(2 mol.L^-1)^2*(3mol.L^-1)
                 = 0.12 mol.L^-1.S^-1
6 0
3 years ago
An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula
uysha [10]

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

  • 0.02 mol Al / 0.02 = 1  Al
  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

8 0
3 years ago
How much carbon dioxide can a single tree absorb in its lifetime?
Bess [88]

Answer:

a tree absorb as much as 48 pound of carbon dioxide per year and can sequester 1 ton of carbon dioxide by the time it reaches 40 years old

5 0
3 years ago
Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate
uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: m_{(AcO)_2Cu} = 0.972 g

Volume of the sodium chromate solution: V_{Na_2CrO_4} = 150.0 mL

Molarity of the sodium chromate solution: c_{Na_2CrO_4} = 0.0400 M

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

n_{(AcO)_2Cu} = 0.0053515 mol

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M

8 0
3 years ago
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