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2 years ago

15

What is a phase change that occurs when a solid transitions directly into a gas state without becoming a liquid?

1 answer:

Sedbober [7]2 years ago

8 0

The phase change is sublimation, so choice A.

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4. What is the difference between an open system and a<br> closed system?

Nimfa-mama [501]

Systems can exist in three ways as open systems, closed systems, and isolated systems. The main difference between open and closed system is that in an open system, matter can be exchanged with the surrounding whereas, in a closed system, matter cannot be exchanged with the surrounding.

btw why not just look it up on google? btw btw brainliest plz

5 0

3 years ago

All of the following are reasons why equations or balanced EXCEPT

Anni [7]

Answer:

monkey

Explanation:

4 0

2 years ago

start out as one kind of rock, but large amounts of pressure and heat change them into a different kind.

Eddi Din [679]

Answer:

Metamorphic

Explanation:

6 0

2 years ago

A gas has a volume of 1.75L at -23C and 150.0kPa. At what temperature would the gas occupy 1.30L at 210.0kPa?

Katen [24]

Answer:

T2 = 260 K

Explanation:

<em>Given data:</em>

P1 = 150.0 k Pa

T1 = (-23+ 273.15) K = 250.15 K

V1 = 1.75 L

P2 = 210.0 kPa

V2 = 1.30 L

<em>To find:</em>

T2 = ?

<em>Formula:</em>

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}$

<em>Calculation:</em>

T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)

T2 = 260 K

8 0

3 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

4 0

2 years ago