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tigry1 [53]
2 years ago
11

Which statement about gases is true? A. They are made up of particles that always move very slowly. B. They are made up of parti

cles that travel in a curved path when in motion. C. They are made up of hard spheres that vibrate quickly in stationary positions. D. They are made up of hard spheres that are in random motion.

Chemistry
2 answers:
rosijanka [135]2 years ago
7 0

Answer: D. They are made up of hard spheres that are in random motion.

Explanation:

A gas is a <u>state of aggregation of matter</u> in which, under certain conditions of temperature and pressure, <u>its molecules interact weakly with each other, without forming molecular bonds</u>, adopting the shape and volume of the container that contains them and tending to separate everything possible because of its <u>high concentration of kinetic energy</u>.

The molecules of a gas are practically <u>free</u> and have the ability to be distributed throughout the space in which they are contained because <u>the gravitational forces and attraction between them are practically negligible</u> compared to the speed at which they move. .

Therefore, gas molecules do not travel specific trajectories or vibrate in a stationary position, instead <u>they move quickly and randomly through the entire space of the container that contains them.</u>

zalisa [80]2 years ago
7 0

Answer:

They are made up of hard spheres that are in random motion.

Explanation:

You might be interested in
Combustion of 8.652 g of a compound containing c, h, o, and n yields 11.088 g of coz, 3.780 g of h2o, and 3.864 g of no2. How ma
Katyanochek1 [597]

Answer:-  C = 3.024 g, H = 0.42 g, N = 1.176 g and O = 4.032 g

Solution:- The compound contains C, H, N and O. On combustion, all the carbon is converted to carbon dioxide, All hydrogen is converted to water and all the nitrogen is converted to nitrogen dioxide.

From the grams of all these, we could calculate their moles and then using mol ratio of these products and the number of moles of C, H and N present in them, we calculate the moles of C, H and N respectively. Further, these moles are converted to grams. On subtracting the sum of grams of C, H and N from the mass of the sample, the mass of oxygen is calculated.

The calculations are as follows:

Calculations for the grams of C:-

11.088gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})(\frac{12gC}{1molC})

= 3.024 g C

Calculations for the grams of H:-

3.780gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})(\frac{1gH}{1molH})

= 0.42 g H

Calculations for the grams of N:-

3.864gNO_2(\frac{1molNO_2}{46gNO_2})(\frac{1molN}{1molNO_2})(\frac{14gN}{1molN})

= 1.176 g N

Mass of the compound is given as 8.652 g. Now we could calculate the grams of oxygen as:

mass of oxygen = 8.652 - (3.024 + 0.42 + 1.176)

= 8.652 - 4.62

= 4.032 g

So, 8.652 grams of the compound contains 3.024 g of C, 0.42 g of H, 1.176 g of N and 4.032 g of O.


7 0
2 years ago
Calculate the height of a column of liquid benzene (d=0.879g/cm3), in meters, required to exert a pressure of 0.790 atm .
Simora [160]

Here we have to get the height of the column in meter, filled with liquid benzene which exerting pressure of 0.790 atm.

The height of the column will be 0.928 m.

We know the relation between pressure and height of a liquid placed in a column is: pressure (P) = Height (h) × density of the liquid (ρ) × gravitational constant (g).

Here the pressure (P) is 0.790 atm,

or [0.790 × (1.013 × 10⁶)] dyne/cm². [As 1 atm is equivalent to 1.013 × 10⁶ dyne/cm²]

Or, 8.002ₓ10⁵ dyne/cm².

density of benzene is given 0.879 g/cm³.

And gravitational constant (g) is 980 cm/sec².

On plugging the values we get:

8.002×10⁵ = h × 0.879 × 980

Or, h = 928.931 cm

Or, h = 9.28 m (As 1 m = 100 cm)

Thus the height will be 9.28 m.

7 0
2 years ago
Read 2 more answers
Magnesium reacts with iron(III) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants
svet-max [94.6K]

Answer:

154.0831 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Mg

Given mass = 41.0 g

Molar mass of Mg = 24.31 g/mol

Moles of Mg = 41.0 g / 24.31 g/mol = 1.6865 moles

Given: For FeCl_3

Given mass = 175 g

Molar mass of FeCl_3 = 162.2 g/mol

Moles of FeCl_3 = 175 g / 162.2 g/mol = 1.0789 moles

According to the given reaction:

3Mg_{(s)}+2FeCl_3_{(s)}\rightarrow 3MgCl_2_{(s)}+2Fe_{(s)}

3 moles of Mg react with 2 moles of FeCl_3

1 mole of Mg react with 2/3 moles of FeCl_3

1.6865 mole of Mg react with (2/3)*1.6865 moles of FeCl_3

Moles of FeCl_3 = 1.1243 moles

Available moles of FeCl_3 = 1.0789 moles

Limiting reagent is the one which is present in small amount. Thus, FeCl_3 is limiting reagent. (1.0789 < 1.1243)

The formation of the product is governed by the limiting reagent. So,

2 moles of FeCl_3 gives 3 moles of magnesium chloride

1 mole of FeCl_3 gives 3/2 moles of magnesium chloride

1.0789 mole of FeCl_3 gives (3/2)*1.0789 moles of magnesium chloride

Moles of magnesium chloride = 1.61835 moles

Molar mass of magnesium chloride = 95.21 g/mol

Mass of magnesium chloride = Moles × Molar mass = 1.61835 × 95.21 g = 154.0831 g

6 0
2 years ago
Convert 1.21 kg to grams
MissTica

Answer:

1210 grams

Explanation:

Message me for extra info.

snap- parkguy786

7 0
2 years ago
Read 2 more answers
A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c
Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0
2 years ago
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