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shtirl [24]
3 years ago
15

What 2 factors increase the validity of a scientific experiment?

Chemistry
2 answers:
zalisa [80]3 years ago
8 0

Answer:

You can increase the validity of an experiment by controlling more variables, improving measurement technique, increasing randomization to reduce sample bias, blinding the experiment, and adding control or placebo groups.

Explanation:

Nimfa-mama [501]3 years ago
7 0

Answer:

Controlling more variables and improving measurement technique.

Btw, I like your profile picture lol

You might be interested in
Calculate the pHpH of a 0.10 MM solution of barium hydroxide, Ba(OH)2Ba(OH)2. Express your answer numerically using two decimal
Maurinko [17]

Answer:

13.301

Explanation:

To calculate the pH of the solution, we must obtain the pOH of the solution as illustrated below:

The dissociation equation is given below

Ba(OH)2 <==> Ba^2+ + 2OH^-

Since Ba(OH)2 dissociate to produce 2moles of OH^-, the concentration of OH^- = 2x0.1 = 0.2M

pOH = - Log[OH^-]

pOH = - Log 0.2

pOH = 0.699

But

pH + pOH = 14

pH = 14 — pOH

pH = 14 — 0.699

pH = 13.301

6 0
3 years ago
Read 2 more answers
Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
Помогите пожалуйста или с первым вопросом или со вторым. Буду очень благодарна, т.к. это мой зачет... :С
Bess [88]
Hsiaqoanwbwiso iDisks
5 0
3 years ago
Fire extinguishers that spray carbon dioxide on the fire, work very effectively because it forms a blanket around the burning ma
Arte-miy333 [17]

Answer:

I think it will option B it will retain enough heat

8 0
3 years ago
What is the volume of 60 g of ether if the density of ether is 0.70g/mL
miss Akunina [59]

Answer:

114mL.

Explanation: hope this helped

8 0
3 years ago
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