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3 years ago

In a controlled scientific experiment,a scientist

Chemistry

2 answers:

zmey [24]3 years ago

8 0

Answer: D

Explanation:

In a controlled scientific experiment, there should be an independent variable, a dependent variable, and a control. The independent variable is the factor in the experiment that is manipulated by the scientist. The dependent variable is the factor in the experiment that changes in response to the independent variable. Controls in an experiment are used as comparison factors, and they can help determine the magnitude of the experiment's results.

Scientists carefully control their variables so they can understand how the variables affect the system as a whole. This makes the data collected from an experiment meaningful.

s344n2d4d5 [400]3 years ago

5 0

The answer is D, alters the test variable, and observes the effects on other outcome variables.

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A 2.0 mL sample of air in a syringe exerts a pressure of 1.02 atm at 22 C. If that syringe is placed into boiling water at 100 C

emmainna [20.7K]

Answer: The new volume of the air in the syringe is 7.3 mL.

Explanation:

Given: $V_{1}$ = 2.0 mL, $P_{1}$ = 1.02 atm, $T_{1} = 22^{o}C$

$V_{2}$ = ?, $P_{2}$ = 1.27 atm, $T_{2} = 100^{o}C$

Formula used to calculate the new volume in syringe is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1.02 atm \times 2.0 mL}{22^{o}C} = \frac{1.27 atm \times V_{1}}{100^{o}C}\\V_{1} = 7.3 mL$

Thus, we can conclude that the new volume of the air in the syringe is 7.3 mL.

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3 years ago

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Wewaii [24]

Answer:

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3 years ago

What is 0.000028474 in scientific notation

antiseptic1488 [7]

Answer:

$2.8474 \: \times \: {10}^{ - 5}$

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3 years ago

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RoseWind [281]

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A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )

Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = $\frac{0.1200}{1000}\times 50.0$ moles = 0.00600 moles

Neutralization reaction (back titration): $NaOH+HCl\rightarrow NaCl+H_{2}O$

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = $\frac{0.0980}{1000}\times 23.60$ moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

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3 years ago