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3 years ago

What is the second step in the classification of living things?

Chemistry

2 answers:

frozen [14]3 years ago

8 0

Answer:

B.) Assigning them to any of the 3 domains

Explanation:

The second step in the classification of living beings is to assign them to any of the three domains. There are not six domains. Furthermore, there are neither six nor three kingdoms, there are 5 kingdoms.

The three domains are:

→ Bacteria Domain: It groups most of the unicellular prokaryotes, such as bacteria and cyanobacteria.

→ Domain Archaea: It groups prokaryotes with different biochemical characteristics and that usually inhabit regions with extreme conditions.

→ Eukaria domain: It groups all unicellular and multicellular eukaryotic organisms, such as fungi, animals, plants and organisms previously classified as protists.

Leya [2.2K]3 years ago

6 0

I'll go with B, have a good day

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Which of the following is NOT<br> necessary to define a coordinate<br> system?

gulaghasi [49]

Answer:

i think that its A,C,D

Explanation:

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2 years ago

NaCl and AgNO3 react in a double replacement reaction. Which is one of the products of this reaction?

hoa [83]

Answer:

nano3+agcl2

Explanation:

double displacement reaction

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3 years ago

Stoichiometry is best defined as the

Oxana [17]

Stoichiometry is “quantitative relationship” among the “reactants” and the “products” in a “chemical reaction”.

<u>Explanation</u>:

In stoichiometry “stoicheion” means element and “metron” means measure in Greek. The stoichiometric calculation depends upon “stoichiometric coefficients” in a “chemical equation” which can be explained as the “number of moles” of each substance (reactants or products). Stoichiometric calculation is done as follows:

For example reaction between nitrogen and hydrogen to form ammonia as

$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$

Here stoichiometric coefficients show that “one molecule of nitrogen” reacts with “three molecules of hydrogen” to form “two molecules of ammonia”. Multiplying Avogadro number $6.022 \times 10^{23}$ to no of molecules in equation:

$\text { For } \mathrm{N}_{2}: 1 \times 6.022 \times 10^{23}=1 \mathrm{mol}$

$\text { For } 3 \mathrm{H}_{2}: 3 \times 6.022 \times 10^{23}=3 \mathrm{mol}$

$\text { For } 2 \mathrm{NH}_{3}: 2 \times 6.022 \times 10^{23}=2 \mathrm{mol}$

Taking molar masses into consideration:

$\text { For } \mathrm{N}_{2}: 1 \times 28.0=28.0 \mathrm{gm}$

$\text { For } 3 \mathrm{H}_{2}: 3 \times 2=6.0 \mathrm{gm}$

$\text { For } 2 \mathrm{NH}_{3}: 2 \times 17=34 \mathrm{gm}$

Hence balanced equation gives stoichiometric coefficients which gives proportion by moles.

8 0

3 years ago

which of these pieces of equipment would be the most appropriate for precisely measuring 29 mL of liquid? Explain your reasoning

Anna007 [38]

Answer:

The best equipment would be the graduated cylinder. Why?

Firstly, the smallest marking on the graduated cylinder is 2 mL, while on all the others the smallest marking is way above that, like 25 mL and 100 mL.

Without even going into the details, we can first rule out the volumetric flask, since its smallest marking is 100 mL and even that is already bigger than our sample size, hence we would have no markings to accurately measure out 29 mL of our sample had we used the volumetric flask.

Next to be ruled out would be the Erlenmeyer flask, as you can see in the image, it only has three marking, and as the smallest marking is 25 mL, each marking is at least 25 mL, and even so far as going up to 50 mL. This cannot let us accurately measure 29 mL out at all, due to the markings being way too big to do that. Hence, the Erlenmeyer flask is ruled out.

Finally, the beaker seems to be a worthy candidate! Unfortunately, for the same reason as the Erlenmeyer flask, as you can see in the image each marking represents 10 mL. We cannot measure 9 mL in the beaker accurately, and hence the beaker is ruled too.

We are left with the graduated cylinder, and that is our answer.

Explanation:

Hope this helped!

5 0

3 years ago

What's the ionic compound formula for zinc bicarbonate

raketka [301]

Zn(HCO3)2 is the compound for Zinc Bicarbonate

6 0

3 years ago

Read 2 more answers