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yKpoI14uk [10]
3 years ago
5

What is carbon dioxide(co2).

Chemistry
2 answers:
olchik [2.2K]3 years ago
8 0
It is a colorless gas.  It is a by product of respiration.
a_sh-v [17]3 years ago
8 0
By a product of respiration, it has no color but is a gas that sometimes isnt seen.
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The smallest particle of an element is a(n) .<br>atom<br>molecule<br>DNA​
mamaluj [8]

Answer:

An atom is the smallest particle of an element

5 0
3 years ago
One solution turns blue. A possible hydrogen ion
aalyn [17]

Answer:

1x10^-8 M

Explanation:

Since the solution turns blue, it mean the solution is a base.

Now, to know which option is correct, we need to determine the pH of each solution. This is illustrated below:

1. Concentration of Hydrogen ion, [H+] = 1x10^-2 M

pH =..?

pH = - log [H+]

pH = - log 1x10^-2

pH = 2

2. Concentration of Hydrogen ion, [H+] = 5x10-2 M

pH =..?

pH = - log [H+]

pH = - log 5x10^-2

pH = 1.3

3. Concentration of Hydrogen ion, [H+] = 5x10 M

pH =..?

pH = - log [H+]

pH = - log 5x10

pH = - 1.7

4. Concentration of Hydrogen ion, [H+] = 1x10-8 M

pH =..?

pH = - log [H+]

pH = - log 1x10^-8

pH = 8

A pH reading shows if the solution is acidic or basic. A pH reading between 0 and 6 indicates an acidic solution, a pH reading of 7 indicates a neutral solution while a pH reading between 8 and 14 indicates a basic solution.

From the above calculations, the pH reading indicates a basic solution when the hydrogen ion concentration was 1x10^-8 M.

5 0
3 years ago
Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -&gt; 2Fe(s) + 3CO(g)
Gelneren [198K]

Answer:

ΔS = +541.3Jmol⁻¹K⁻¹

Explanation:

Given parameters:

Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹

Standard Entropy of C = 5.7Jmol⁻¹K⁻¹

Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹

Standard Entropy of CO = 198Jmol⁻¹K⁻¹

To find the entropy change of the reaction, we first write a balanced reaction equation:

                              Fe₂O₃ +  3C →  2Fe + 3CO

To calculate the entropy change of the reaction we simply use the equation below:

      ΔS = ∑S_{products} - ∑S_{reactants}

Therefore:

     ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1

                          ΔS = +541.3Jmol⁻¹K⁻¹

5 0
3 years ago
For the main group elements( excluding the transition elements) , is it necessary to memorize the type of ion each element makes
vagabundo [1.1K]
Yes, you should memorize them. While it is easy to predict them by looking at a periodic table, you should still have a strong grasp of simpler chemistry such as this before more challenging topics are taught.
3 0
3 years ago
How many moles of FeCI3 are present in 345.0 g FeCI3
Sphinxa [80]

Answer:

\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

  • Fe:  55.84 g/mol
  • Cl:  35.45 g/mol

There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

  • FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol

This number tells us the grams of FeCl₃ in 1 mole.

2. Calculate Moles

Use the number as a ratio.

\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Multiply by the given number of grams, 345.0.

345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Flip the fraction so the grams of FeCl₃ will cancel.

345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}

345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }

\frac{345.0 \ mol \ FeCl_3}{162.19 }

Divide.

2.12713484 \ mol \ FeCl_3

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

\approx 2.127 \ mol \ FeCl_3

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

6 0
3 years ago
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