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ss7ja [257]
3 years ago
11

How do solids, liquids, and gases differ? in solid matter, atoms or molecules pack close to each other in fixed locations; in ga

ses, atoms or molecules pack about as closely as they do in solid matter, but they are free to move; in liquid matter, atoms or molecules have a lot of space between them and are free to move a solid has a fixed volume and rigid shape; liquids assume the shape of their container; gases always assume the shape and volume of their container in solid matter, atoms or molecules pack close to each other but, they are free to move; in liquid matter, atoms or molecules pack about as closely as they do in solid matter, they are also free to move; in gaseous matter, atoms or molecules have a lot of space between them and are free to move relative to one another a solid has a fixed volume and rigid shape; gases assume the shape of their container; liquids always assume the shape and volume of their container
Chemistry
1 answer:
geniusboy [140]3 years ago
5 0

in solid matter, atoms or molecules pack close to each other in fixed locations; in gases, atoms or molecules pack about as closely as they do in solid matter, but they are free to move;

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kvasek [131]

election circles the nucleus

6 0
3 years ago
The absorbance of a garbanzo bean solution that had been diluted by a factor of three was 0.528. what was the concentration of t
Serjik [45]
The Beer-Lambert law states that A = E*c*l where A is absorbance, E is the molar absorbance coeffecient, c is concentration and l is path length. Therefore the absorbance is directly proportional to concentration, and by increasing the concentration by a factor of 3, absorbance will increase by a factor of 3 giving   A = 1.584
6 0
4 years ago
If 1. 3618 moles of AsF3 are allowed to react with 1. 0000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles
VARVARA [1.3K]

Answer:

AsF3:C2CI6

4:3

1.3618 moles: 1.02135 moles(1.3618÷4×3)

C2CI6 is the limting reagent

So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)

or

Balanced equation

4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4

Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.

Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.

Explanation:

5 0
2 years ago
What do dwarf planets, asteroids, and comets have in common
adell [148]

Answer:

their all small and arent considered planets

hope this could help

Explanation:

6 0
3 years ago
Read 2 more answers
Write a balanced equation for the complete combustion of each compound. a. formaldehyde (CH2O(g)) b. heptane (C7H17(l)) c. benze
Yuri [45]

Answer:

a. formaldehyde (CH₂O(g)): CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

b. heptane (C₇H₁₇(l)): 4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l))​: C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

Explanation:

In a reaction of combustion, a hydrocarbon compound (composed of C, H and O) reacts with oxygen gas (O₂). The <u>complete</u> combustion of a hydrocarbon - such as formaldehyde, heptane, and benzene - produces carbon dioxide (CO₂) and water (H₂O).  

Thus, we write the reactants and products for each combustion reaction and then we balance the atoms: C, H, and O.

a. formaldehyde (CH₂O(g)):

CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

The chemical equation is already balanced with the coefficients 1: we have the same number of C atoms (1), H (2) and O (3) on both sides of the equation.

b. heptane (C₇H₁₇(l)):

C₇H₁₇(l) + O₂(g) → CO₂(g) + H₂O(g)

Here we have to write a coefficient 28 for CO₂ to balance the C atoms in the products side, and for C₇H₁₇ we write 28/4 = 7. With similar reasoning we found the coefficients for O₂ and H₂O:

4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l))​:

C₆H₆(l) +  O₂(g) → CO₂(g) +  H₂O(g)

First, we write a coefficient of 6 for CO₂ to balance the C atoms. Then, we have to balance H atoms: we write a coefficient 3 in H₂O. Now, we have 12 + 3 = 15 atoms of O on the reactants side. So, we write a half of these number of atoms in the coefficient for O₂: 15/2. We obtain the balanced equation:

C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

5 0
3 years ago
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