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dolphi86 [110]
3 years ago
13

Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter freq

uencies than visible light?
1. Gamma rays
2. Infrared radiation
3. Ultraviolet liht
4. X-rays
5. Microwaves
6. Radio waves
Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
5 0

Answer:

3,4,1 and 6,5,2

Explanation:

In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;

Radio waves

Microwaves

Infrared waves

Visible light rays

Ultraviolet rays

X-rays

Gamma rays

(a simple mnemonic is RMIVUXG)

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Differentiate between the various qualities of reagents which are ordinary ,analar and industrial quality reagent
dolphi86 [110]
The difference between ordinary, analar and industrial reagents is the in the purity. Ordinary reagents are those whose purity meets the standard. Analar reagents are used in chemical analysis and they are of high purity but with known contaminants which again illustrates their use in chemical analyses.Finally the industrial reagents are not pure enough and they are used for industrial purposes as well for commercial use.
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How many atoms are in an Apple
miss Akunina [59]

Answer:10^27 atoms in an apple

Explanation:

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3 years ago
Determine the empirical formula for the compound represented by each molecular formula. c2h4
dybincka [34]
Carbon and hydrogen and the number next to is the number of moles
6 0
4 years ago
If you weighed the atoms that appear on the reactant side of the equation, would they have the same mass as the atoms that appea
tatuchka [14]

Answer:

No

Explanation:

If you added the reactants on the reactant side because there is one atom for nitrogen on the product side while there is two atoms on the product side. There are more hydrogen products on the reactant side as well.

3 0
3 years ago
3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15
shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}

(b) Mass of H

\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0
4 years ago
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