All categories

2 years ago

The three main types of scientific investigations are descriptive, comparative, and experimental. Which components are

2 answers:

7 0

The components included in all three are a scientific question, a procedure, and a conclusion. A scientific question is used to identify the purpose of an experiment, which an investigation can be designed from. A procedure must be present in every experiment so that there is structure to every investigation. Conclusions are always present, because they summarize the results and outcomes of the experiment.

Marina86 [1]2 years ago

5 0

Answer:the abswe is a,d, and f

Explanation: the peson below couldve just said that thx have a good day:)

You might be interested in

An industrial chemist is studying a sample of an unknown metal. Describe two ways he could change the metal physically and two w

Illusion [34]

Answer:idk what they cpuld do but we can say this is the right answer :))))

Explanation:

I need PoInTs

5 0

3 years ago

What is the length of time an organism is expected to live?

Nataly [62]

This is called a life span.

5 0

3 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol =<u> 12.81 grams</u>

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = <u>75.70 grams CO2</u>

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = <u>34.29 grams</u>

3 0

3 years ago

What substance would you add to KF(aq) to form a buffer solution? View Available Hint(s) What substance would you add to to form

AnnyKZ [126]

<h3>Answer</h3>

<h3>Explanation</h3>

A buffer solution contains <em>a weak acid</em> and<em> its conjugate base</em>. The two species shall have a similar concentration in the solution. It's also possible for <em>a weak base</em> and <em>its conjugate acid</em> to form a buffer solution.

The KF solution already contains large number of $\text{F}^{-}$ ions. The objective is to thus find its conjugate acid or base.

$\text{F}^{-}$ contains no proton $\text{H}^{+}$ and is unlikely to be a conjugate acid. Assuming that $\text{F}^{-}$ is a conjugate base. Adding one proton to $\text{F}^{-}$ would produce its conjugate acid.

$\text{H}^{+} \; (aq) + \text{F}^{-} \; (aq) \rightleftharpoons \text{HF}\; (aq)$

Therefore $\text{HF}$ is the conjugate acid of $\text{F}^{-}$ . $\text{HF}$ happens to be a weak acid. As a result, combining $\text{F}^{-}$ with $\text{HF}$ would produce a solution with large number of both the weak acid and its conjugate base, which is a buffer solution by definition.

4 0

3 years ago

ILL GIVE A BRAINLY PLS HELP

Oksana_A [137]

Atoms are the unit of the molecule of the compound. The 3.01 x 10²⁴ atoms of oxygen are present in 5 moles of water and 3 moles of carbon dioxide in the sample.

<h3>What are atoms?</h3>

Atoms are the smallest fundamental unit of the compounds that can be given by Avogadro's number.

For calculating the oxygen atoms in 5 mole water:

If 1 mole = 6.02 × 10²³

Then, 5 moles = 5 × 6.02 × 10²³

Hence, 3.01 x 10²⁴ atoms of oxygen are present in 5 moles of water.

Moles of carbon dioxide in the sample is calculated as:

If 1 mole of carbon dioxide = 6.02 × 10²³ molecules

Then moles in 1.8 x 10²⁴ molecules will be,

(1.8 x 10²⁴ molecules) ÷ (6.02 × 10²³ molecules) = 3 moles

Hence, 3 moles of carbon dioxide is present in the sample.

Learn more about Avogadro's number here:

brainly.com/question/5098076

#SPJ1

3 0

2 years ago