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3 years ago

To which wave behavior can audible beats be attributed?

1 answer:

8 0

Audible beats can be attributed to :
A. constructive interference.
Constructive interference happened when two waves meet in such a way that their crests lined up together.

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Copper atoms are oxidised at the positive electrode to Cu2+ ions, as shown in the half equation.

mafiozo [28]

(anode +) Cu⁰(s) - 2e⁻ → Cu²⁺(aq)

3 0

4 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.mg of oxalic acid H2C2

Nezavi [6.7K]

Answer:

The molarity of the student's sodium hydroxide solution is 0.0219 M

Explanation:

1) Chemical reaction.

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

2) Mole ratio

2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

3) Starting amount of oxalic acid

mass = 28 mg = 0.028 g
molar mass = 90.03 g/mol
Convert mass in grams to number of moles, n:

n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol

4) Titration

Volume of base: 28.4 mL = 0.0248 liter
Concentration of base: x (unknwon)

Number of moles of acid: 2.52 mol (calculated above)

Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

$\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622$

That means that there are 0.000622 moles of NaOH (solute)

5) Molarity of NaOH solution

M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using three signficant figures, such as the starting data are reported.

5 0

3 years ago

Read 2 more answers

Select all that apply to organic compounds. Organic compounds are only found in living things. Organic compounds can be synthesi

Andreyy89

Answer:

Correct options:

Organic compounds can be synthesized in a laboratory setting.

There are hundreds of thousands of organic compounds.

Explanation:

1) Organic compounds are only found in living things: False

Organic compounds are mainly carbon based. Although organic compounds are derived from living things like plants and animals, non-living things like sedimentary rocks are also rich in carbon.

2) Organic compounds can be synthesized in a laboratory setting: True

There are four major categories of organic compounds: carbohydrates, proteins, lipids and nucleic acids all of which can be synthesized in a laboratory.

3) Organic compounds contain carbon: False

Organic compounds are identified by the presence of both carbon and hydrogen (and sometimes other elements like O, N, S).

4)CO2 is an organic compound: False

There are no H atoms, hence CO2 is not organic.

5) There are hundreds of thousands of organic compounds: True

7 0

3 years ago

Read 2 more answers

The addition of nutrients to water by human activity is called artificial

nikitadnepr [17]

Answer:

The addition of nutrients to water by human activity is called artificial Eutrophication. Nonpoint-source-pollution comes from many small sources.

Explanation:

hope this helps :)

5 0

4 years ago

How much heat is needed to raise the temperature of 50.0 g of water from 4.5 ⁰C to 83.0 ⁰C?

Alex_Xolod [135]

The heat needed to raise the temperature of 50.0 g of water from 4.5 ⁰C

to 83.0 ⁰C is 16.46KJ

Explanation:

To solve the problem,

we need the value of the water and its given by,

C=4.18 × J/ g ∘ C

To increase the temperature we need m grams of water by n ∘ C

And the formula for the heat is given by, heat= m × n ×specific

heat

q = m ⋅ c . Δ T

where q denotes the heat absorbed

m denotes the mass sample

c denotes the specific heat

Δ T denotes the change in the temperature

Based on the data it is written as,

q= 50.0 g . 4.18 J / g ∘ C (83.0-4.5) ∘ C

Simplifying the equation we get,

q=16406.5 J

Express it in the form of significant figure

16406.5 J . 1 KJ / 10^ 3 J

Solving the equation we get

16.46KJ

The heat needed to raise the temperature of 50.0 g of water from 4.5 ⁰C

to 83.0 ⁰C is 16.46 KJ

6 0

3 years ago