A certain mineral was able to catch on fire. The ability to catch on fire is a ___________

write down the way you had to behave to illustrate the behavior of particles as a gas condenses to form a liquid

Rainbow [258]

Gases near together and vibrate in position however, don't circulate beyond each other. In a liquid, the particles are interested in every different but now not as a great deal as they may be in a strong.

The particles of a liquid are near together, constantly transferring, and may slide beyond one another. The Kinetic-molecular concept attempts to explain the behavior of fuel molecules based totally on the nature of gasoline. The principle is grounded on simple assumptions

In gases the debris passes swiftly in all directions, regularly colliding with every different facet of the box. With a boom in temperature, the debris gains kinetic strength and passes more quickly. Gasoline is a state of matter that has no constant form and no fixed extent. Gases have a decreased density than other states of the count, together with solids and liquids. there may be a high-quality deal of empty area between debris, that have loads of kinetic energy and aren't especially drawn to one another.

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3 0

1 year ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1

H = $\frac{0.1298}{0.0652}$ = 2

O = $\frac{0.130}{0.0652}$ = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is $CH_2O_2$ .

3 0

3 years ago

A balance and a graduated cylinder are used to determine the density of a liquid sample. The sample has a mass of 14.0 g and a v

mylen [45]

Answer:

Density = 7 g/mL

Explanation:

Given that,

Mass of a sample, m = 14 g

Volume of the sample, V = 2 mL

We need to find the density of the liquid sample. We know that the density of an object is given by :

$d=\dfrac{m}{V}\\\\d=\dfrac{14\ g}{2\ mL}\\\\d=7\ g/mL$

So, the density of the liquid sample is 7 g/mL.

7 0

3 years ago

Potassium hydroxide molar mass

Mnenie [13.5K]

The answer I will give is an approximate number, but it will be close. The problem is that every periodic table is slightly different. Some round as I will, and some carry the mass of the elements out to 3 decimal places. You should go back and put the correct numbers in to the mass that I use.

Formula

KOH

Givens

K = 39
O = 16
H = 1

Solution

Molar Mass = 39 + 16 + 1

Molar Mass = 56 grams / mole

6 0

3 years ago

What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?

ioda

N₂O is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen.

Empirical formula can be calculated by

Suppose we have 100 g of the substance. That indicates that it has 36.69 grams of oxygen and 63.61 grams of nitrogen.

Masses transformed into moles:

Formula used

Given mass/ Molar mass

14.01 g contains 1 mol of N

So 63.61 g of N contains moles is equals to

(1 mol N / 14.01 g N) 63.61 g N = 4.540 mol N

Similarly

16 g of O contains 1 mole of O

36.69 g of O contains moles is equals to

(1 mol O / 16.00 g O) 36.69 g O = 2.293 mol O

Divide by the smallest to normalize:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

Therefore, there are roughly twice as many N as O atoms. N2O is the empirical formula as a result.

Ratio is basically 2:1

Hence, N₂O is the empirical formula of an oxide of nitrogen

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7 0

2 years ago

A certain mineral was able to catch on fire. The ability to catch on fire is a ____________ property