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1 year ago

15

Which type of electromagnetic wave has less energy than a microwave?

1 answer:

polet [3.4K]1 year ago

7 0

Radio waves.

From lowest to highest it is radio wave, microwave, infrared, visible light, ultraviolet, x ray, and then gamma.

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Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of

dsp73

The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.

Molarity is mol HNO3 / L of solution. This is our aim

The given percentage is 0.68 g HNO3/ g solution

multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain

0.016 mol/ mL or 16.23 mol/ L (M)

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2 years ago

What is the sign of the charge of an electron

oksian1 [2.3K]

Answer:

Electrons have negative charge.

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2 years ago

Classify the type of this chemical reaction: <br> C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

sweet-ann [11.9K]

The given reaction is a combustion reaction, since a hydrocarbon is burning in presence of oxygen

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2 years ago

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Sodium chloride (NaCl) is an ionic solid. It dissolves easily in water. Which property of water allows it to dissolve NaCl?

WINSTONCH [101]

Its hydrogen atom is a high conductor of electricity

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2 years ago

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

<u />

<u>4. Equilibrium expression</u>

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

<u />

<u>5. Solve:</u>

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0

2 years ago