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Novosadov [1.4K]
3 years ago
7

Describe the atoms in matter as it changes from a gas to a liquid. ANSWER QUICK!! 90 POINTS IF ANSWER

Chemistry
1 answer:
Sonja [21]3 years ago
3 0

Answer:

Clouds are water

Explanation:

When a gas gathers up into a cloud,the cloud starts to rain and water comes out of the cloud.

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The answer is E. coli
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In an acid-base titration,
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b

Explanation:

An acid-base titration is an experimental procedure used to determined the unknown concentration of an acid or base by precisely neutralizing it with an acid or base of known concentration. ... It is filled with a solution of strong acid (or base) of known concentration.

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When two species A and B form an electron-pair bond and A does not provide its electrons for bonding, the bond present between A
vichka [17]

Answer:

Ionic

Explanation:

If A does not have electron to bond, it just receives one electron from B.

It can´t be covalent because A don´t have any electrons to bond with B.

5 0
3 years ago
Activity
Monica [59]

Answer:

The mass of a pure substance represents a specific number of pure substance 'molecules', as defined by molecular weight. The relative atomic mass (atomic weight) in grams/mole (g/mol) is the molar mass of a substance, which means it is the mass of one mole of that substance.

Explanation:

Hope this helps :)

8 0
4 years ago
A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
3 years ago
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