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2 years ago

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In an experiment, what is an independent variable?

1 answer:

Andrews [41]2 years ago

8 0

Answer:

The independent variable (IV) is the characteristic of a psychology experiment that is manipulated or changed by researchers, not by other variables in the experiment.

For example, in an experiment about the effect of nutrients on crop growth: The independent variable is the amount of nutrients added to the crop field. The dependent variable is the biomass of the crops at harvest time.

Therefore : 3. the variable that responds to what the scientist changes in an experiment

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Climate change as we know it today is

USPshnik [31]

The answer is a!!!!!!!!!!

7 0

2 years ago

Identify the conjugate base in each pairs

DiKsa [7]

Answer: 1) $RCOO^-$

2) $H_2PO_4^-$

3) $RNH_2$

4) $HCO_3^-$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

1) $RCOOH\rightarrow RCOO^-+H^+$

Here, $RCOOH$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $RCOO^-$ which is a conjugate base.

2) $H_3PO_4\rightarrow H_2PO_4^-+H^+$

Here, $H_3PO_4$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $H_2PO_4^-$ which is a conjugate base.

3) $RNH_3^+\rightarrow RNH_2+H^+$

Here, $RNH_3^+$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $RNH_2$ which is a conjugate base.

4) $H_2CO_3\rightarrow HCO_3^-+H^+$

Here, $H_2CO_3$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $HCO_3^-$ which is a conjugate base.

8 0

2 years ago

Which two substances are reactants in the following chemical equation?

krok68 [10]

Both A and D
Because these compunds of reactant and the compounds of the product . The reactant goint to equation

4 0

1 year ago

MissTica

Answer:

what is it all about???

8 0

3 years ago

How many moles are in 83.1g of CaBr2

9966 [12]

Answer:

0.416 mol CaBr₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

83.1 g CaBr₂

<u>Step 2: Identify Conversions</u>

Molar Mass of Ca - 40.08 g/mol

Molar mass of Br - 79.90 g/mol

Molar Mass of CaBr₂ - 40.08 + 2(79.90) = 199.88 g/mol

<u>Step 3: Convert</u>

<u /> $83.1 \ g \ CaBr_2(\frac{1 \ mol \ CaBr_2}{199.88 \ g \ CaBr_2} )$ = 0.415749 mol CaBr₂

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.415749 mol CaBr₂ ≈ 0.416 mol CaBr₂

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2 years ago