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2 years ago

Defention of density in your own words

Chemistry

2 answers:

Readme [11.4K]2 years ago

8 0

Answer:

Density is a word we use to describe how much space an object or substance takes up (its volume) in relation to the amount of matter in that object or substance (its mass). Another way to put it is that density is the amount of mass per unit of volume. If an object is heavy and compact, it has a high density.

Explanation:

Burka [1]2 years ago

7 0

Answer:

density: the ratio of mass to volume for a substance

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In this experiment you’ll be grabbing the relationship between mass and volume you’ll be placing the volume on the X axis and th

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The mass and volume has direct proportional to each other.

<h3>What is the relationship between mass and volume?</h3>

The volume of the object is directly proportional to its mass which means that if the volume increases the mass of the object increases also increase and vice versa. Due to this direct relationship, the line of regression will go straight between X and Y axis.

So we can conclude that mass and volume has direct proportional to each other.

Learn more about volume here: brainly.com/question/1972490

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3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

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36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$