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3 years ago

A substance that forms a vapor is generally in what physical state at room temperature?liquid or solidsolidliquidgas

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

6 0

The answer is liquid or solid.

That is a substance that forms a vapor is generally in liquid or solid physical state.

The substance that is in liquid or solid physical state, forms vapor.

When heated solid and liquid can be converted into vapor so the answer is solid or liquid is the physical state that forms vapors.

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Chromium (VI) forms two different oxyanions, the orange dichromate ion, Cr2o72-, And the yellow chromate ion, CrO4 2-. The equil

MrRissso [65]

Answer:

See explanation

Explanation:

Let us look at the reaction again;

Cr2O7 2- (aq) + H2O(l)⇄ 2CrO4 2-(aq) + 2H^+(aq)

When we add sodium hydroxide to the system as shown, the hydroxide ion removes the hydrogen ion thereby leaving a large concentration CrO4^2-(aq) in the system this causes the solution to turn green(equilibrium position shifts to the right).

The net ionic equation is;

OH^-(aq) + H^+(aq) ----> H2O(l)

The reaction;

OH^-(aq) + H^+(aq) ----> H2O(l) is exothermic hence, if the temperature of the system is increased, the equilibrium position will shift towards the left hand side and the solution turns orange.

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2 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago

What is the concentration of the unknown h3po4 solution? the neutralization reaction ish3po4(aq)+3naoh(aq)→3h2o(l)+na3po4(aq) -g

hjlf

First, we need to get moles of NaOH:

when moles NaOH = volume * molarity

= 0.02573L * 0.11 M

= 0.0028 moles

from the reaction equation:

H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)

we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH

∴ X mol H3PO4 reacts with → 0.0028 moles NaOH

∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol

now we can get the concentration of H3PO4:

∴[H3PO4] = moles H2PO4 / volume

= 9.4 x 10^-4 / 0.034 L

= 0.028 M

6 0

3 years ago

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Nadusha1986 [10]

Answer:

Explanation:

Runoff occurs when there is more water than land can absorb so it would be C.

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2 years ago

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