Identify the step that is not part of the scientific method.

2 answers:

5 0

State the cause is not part of the scientific method.
the scientific method is as follows:
1) form a hypothesis
2)perform an experiment to test hypothesis
3)draw conclusions from experiment

Lana71 [14]3 years ago

4 0

Answer:

to help u if u don't understand what the other guy said it's D

Explanation:

You might be interested in

Which of the following uses light energy to convert co2 and h2o to organic compounds?

Dimas [21]

Photosynthesis is where carbon dioxide and water through light energy is converted in glucose and other organic compounds.

4 0

4 years ago

A glass holds 6 ounces of 60-proof rum (30 alcohol). how much fruit juice must be added to the rum so that the mixture is dilute

Elenna [48]

We need an equation that will relate the concentrated mixture and the diluted one. To solve this we use the equation,

M1 V1 = M2 V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

30 % x 6 oz = 20 % x V2

V2 = 9 oz

The volume of the diluted mixture would be 9 oz. Therefore, you will need to add 9 oz - 6 oz = 3 oz of fruit juice to dilute the 30 percent alcohol to 20 percent alcohol.

7 0

3 years ago

Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um

Andre45 [30]

Answer:

$\large \boxed{\text{0.0038 mol/L}}$

Explanation:

1. Calculate the initial moles of acid and base

$\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}$

2. Calculate the moles remaining after the reaction

OH⁻ + H₃O⁺ ⟶ 2H₂O

I/mol: 0.0053 0.005 00

C/mol: -0.00500 -0.005 00

E/mol: 0.0003 0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

$\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}$