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3 years ago

What were the four elements that ancient Greeks believed to exist

Chemistry

1 answer:

ruslelena [56]3 years ago

4 0

Answer:

The ancient Greeks believed in the four elements of air, earth, fire, and water. The Greek philosopher Empedocles is known for originating the theory of these classic four elements being the basis of all things.

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An atom of hydrogen and an atom of carbon are

LenKa [72]

Answer: -

Organic compounds consisting of only carbon atoms and hydrogen atoms are known as hydrocarbons.

Hydrocarbons are used by us for mainly as a source of energy. Hydrocarbons make up gasoline which is used as fuel in automobiles.

Hydrocarbons depending on the type of unsaturation can be of three types - alkanes, alkenes and alkynes.

5 0

3 years ago

1.Which is the largest atom in Group 1?

Serhud [2]

1. Francium
2. Nitrogen
3.rubidium
4. Krypton

7 0

3 years ago

Gamma rays are negatively charged.

asambeis [7]

Answer:

True

Explanation:

8 0

3 years ago

Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation y

sammy [17]

Answer:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

Explanation:

We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

5 0

3 years ago

An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate th

Anettt [7]

Answer:

The method is accurate in the calculation of the $Cu^+2$

Explanation:

As a first step we have to calculate the <u>average concentration </u>of $Cu^+2$ find it by the method.

$\frac{0.782+0.762+0.825+0.838+0.761 }{5} =0.79 ppm$

Then we have to find the<u> standard deviation:</u>

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_i-\bar{x})^2}=0.0359$

For the confidence interval we have to use the formula:

μ=Average± $\frac{t*s}{\sqrt{n} }$

Where:

t=t student constant with 95 % of confidence and 5 data=2.78

μ= $0.79$ ± $\frac{2.78*0.0359}{\sqrt{5} }$

upper limit: 0.84

lower limit: 0.75

If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.

7 0

3 years ago