<span>Inference because she drew a conclusion based on evidence.</span>
I believe D. because all the organisms that live in a lake can be a bunch of living things that would be considered a population and not a species. I canceled out B. from the get go because, I don’t believe non living organisms can be in a population ?
Please get back to me if I’m right or wrong :)
Answer : The chemical formula of pure substances are different from those of mixtures because all the matter that exists on earth can be classified into two categories namely; pure substances and mixtures.
A pure substance always consists of a single element or compound. Example Oxygen is formed only of oxygen (O) atoms; whereas the table salt which is formed only by sodium chloride (NaCl) molecules is a compound which is made up of pure elements. A pure substance cannot be further divided into its components.
A mixture is made up of different compounds and/or elements. It can be easily separated using any physical method. Example, sand, salt and saw dust mixed together as a mixture. Also, it will not have a definite chemical formula.
Please refer the attachment for better understanding.
The mass of a lead ball with a radius of 0.50mm is 5.9714226× g
The density of a substance is defined as the mass of substance per unit of its volume. Density is represented by the symbol, ρ.
The formula for density is -
Density (ρ) = <u> mass (m) </u>
volume (v)
Given :
Density of lead = 11.4 g/cm³
Radius of lead ball = 0.50 mm = 0.05 cm
Since,
Vsphere = 4/3 π r³
v = <u>4 </u>× 3.14 × (0.05)³
3
v = 4.18 x (1.25 x 10⁻³)
v = 5.225 x 10⁻³
Since,
Density = <u> mass </u>
volume
11.4 g/cm³ = <u> mass </u>
5.225 x 10⁻³
mass = 11.4 x (5.225 x 10⁻³)
mass = 5.9565 x 10⁻² or 0.05955 g
So, the mass of a lead ball with a radius of 0.50mm is 5.9565 x 10⁻² g
To learn more about density,
brainly.com/question/14084745
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Answer:
To make the buffer you must add 1 mole of MES in the liter of the buffer.
Explanation:
To find pH of the buffer of MES you must use Henderson-Hasselbalch equation:
pH = pKa + log [MES⁻] / [MES]
<em>Where [MES⁻] is molar concentration of unprotoned MES and [MES] molar concentration of the acid</em>
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Replacing:
6.2 = 6.2 + log [1M] / [MES]
<em>Molar concentration of [MES⁻] is 1mol / L = 1M</em>
0 = log [1M] / [MES]
1 = [1M] / [MES]
[MES] = 1M
That means to make the buffer you must add 1 mole of MES in the liter of the buffer
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