All categories

3 years ago

Why is a quantitative observation more useful than a non- quantitative one? Which of the following are quantitative?

1 answer:

7 0

A quantitative observation is not necessarily more useful than a non-quantitative one. However, quantitative observations do allow one to ﬁnd trends.

(a), the sun rising is a non-quantitative observation.

(b), knowledge of the numerical relationship between the weight on the Moon and on Earth, is a quantitative observation.

(c), watching ice ﬂoat on water does not involve a measurement; therefore, it must be a qualitative observation.

(d) the fact that we know that the water pump won’t work for depths more than 34 feet makes it quantitative. Again, seeing numbers is a giveaway that it’s a quantitative <span>observation. Quantitative is where you deal with numbers.</span>

You might be interested in

There is nothing that you can do to conserve energy. True or False

Firdavs [7]

I am going to say it is false.

6 0

3 years ago

Read 2 more answers

How is the preodictable arranged by atomic mass, coloms , and rows

bearhunter [10]

answer:an arrangement of elements in columns, based on a set of properties that repeat from row to row

Explanation:

hope this helps love <3

5 0

3 years ago

Read 2 more answers

Which characteristics from the list below cause terrestrial planets to be more likely to have life than Jovian planets?

ddd [48]

Answer:

the Mercury planet is likely to have life but the earth as a lot of oxygen

6 0

3 years ago

What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso

Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

$pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247$

The pH of the buffer can be known as

$pH = pK_{a} + log[\frac{[A-]}{[HA]}}]$

The concentration of $[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\$

Similarly, the concentration of [HA] = $\frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404$

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

$pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236$

So, the pH of the buffer is 6.1236.

5 0

4 years ago

Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L

const2013 [10]

Answer:

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.

We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

m=nmass.solvent=0.288moles0.250kg=1.15m

4 0

3 years ago

Read 2 more answers