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2 years ago

Scientific method steps

Chemistry

2 answers:

valentinak56 [21]2 years ago

8 0

Answer:

ask a question

research

construct basic hypothesis

test and experiment your hypothesis

analyze data & draw a conclusion

communicate results

Explanation:

...

gregori [183]2 years ago

4 0

Answer:

Make an observation.

Ask a question.

Form a hypothesis, or testable explanation.

Make a prediction based on the hypothesis.

Test the prediction.

Iterate: use the results to make new hypotheses or predictions

Explanation:

You might be interested in

a solution of unknown molecular substance is prepared by dissolving 0.50g of the unknown in 8.0g of benzene. the solution freeze

spayn [35]

Answer:

THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.

Explanation:

Mass of the unknown substance = 0.50 g

Freezing point of the solution = 3.9 °C

Freezing point of pure benzene = 5.5 °C

Freezing point dissociation constant Kf = 5.12°C/m

First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.

Change in temperature = 5.5 -3.9 = 1.6 °C

Next is to calculate the number of moles or molarity of the compound that dissolved.

Using the formula:

Δt = i Kf m

Assume i = 1

So,

1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene

x = 1.6 * 0.008 / 5.12

x = 0.0128 / 5.12

x = 0.0025 moles.

Next is to calculate the molar mass using the formula, molarity = mass / molar mass

Molar mass = mass / molarity

Molar mass = 0.50 g /0.0025

Molar mass = 200 g/mol

Hence, the molar mass of the unknown compound is 200 g/mol

6 0

2 years ago

The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie

Deffense [45]

<u>Answer:</u> Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

<u>For ozone:</u>

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

$\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol$

<u>For nitric oxide:</u>

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

$\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol$

For the given chemical equation:

$O_3+NO\rightarrow O_2+NO_2$

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

$0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g$

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

8 0

3 years ago

PLS HELP QUICK ALOTTT OF POINTS

timofeeve [1]

Answer:

$\boxed {\boxed {\sf 0.80 \ mol\ F}}$

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

$\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

$4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

Flip the ratio so the units of atoms of fluorine cancel each other out.

$4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}$

$4.8 \times 10^{23} *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }$

Condense into 1 fraction.

$\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F$

Divide.

$0.7970773829 \ mol \ F$

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

$0.80 \ mol \ F$

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

6 0

2 years ago

In general how are word equations written to describe chemical equations

Alexus [3.1K]

<span>chemicals reacting are written on the left, what is formed is written on the right after the = sign

eg Copper + oxygen = copper oxide.
Hope this helps , mark as brainliest and say thanks THANKS..</span>

5 0

2 years ago

What volume is occupied by 0.102 molmol of helium gas at a pressure of 0.95 atmatm and a temperature of 305 kk ?

zalisa [80]

The volume occupied by 0.102 mole of the helium gas is 2.69 L

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

Number of mole (n) = 0.102 moles
Pressure (P) = 0.95 atm
Temperature (T) = 305 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?

<h3>How to determine the volume </h3>

The volume of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both sides by P

V = nRT / P

V = (0.102 × 0.0821 × 305) / 0.95

V = 2.69 L

Thus, the volume of the gas is 2.69 L

Learn more about ideal gas equation:

brainly.com/question/4147359

#SPJ1

8 0

1 year ago