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3 years ago

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

Chemistry

1 answer:

Sedbober [7]3 years ago

8 0

The phase change is sublimation, so choice A.

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What are the law of 10 in energy Transfer

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When the energy is passed on from one trophic level to another, only 10 percent of the energy is passed on to the next trophic level.

Explanation:

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3 years ago

When the number of electrons in an atom changes, then the atom's_______ change.

AlekseyPX

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It would change the charge of the atom.

Explanation:

Added electrons cause atoms to be negatively charged, lost electrons cause atoms to be positively charged.

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3 years ago

Jane described two substances which participate in photosynthesis.

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3 years ago

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Enter an equation to show how h2po3− can act as a base with hs− acting as an acid. Express your answer as a chemical equation. I

vladimir2022 [97]

The acid - base equation between H2PO3^- and HS^- is H2PO3^- + HS^- ⇄S^- + H3PO3.

An acid is a substance that can donate hydrogen ions while a base is a substance that can accept hydrogen ion. This is the acid base definition according to Brownstead - Lowry.

To show the acid - base relationship between H2PO3^- and HS^-, we have the equation;

H2PO3^- + HS^- ⇄S^- + H3PO3

Learn more about acids and bases: brainly.com/question/10282816

4 0

2 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

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4 years ago