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3 years ago

What happens to the particles during condensation

Chemistry

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

molecus in a gas cool down

Explanation:

the molecules lose heat and energy do they slow down they move closer to other molecules and form a liquid

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What type of decay is illustrated by the equation below?

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Answer:

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Explanation:

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Suppose 650 mL of hydrogen gas are produced through a displacement reaction involving solid iron and sulfuric acid, H2SO4, at ST

scoray [572]

Answer:

4.415 g of FeSO₄

Explanation:

The balance chemical equation for given single replacement reaction is as follow;

Fe + H₂SO₄ → FeSO₄ + H₂

Data Given;

Volume = 650 mL = 0.65 L

Density at STP = 0.08988 g/L

Mass = Density × Volume = 0.08988 g/L × 0.65 L = 0.0584 g

Step 1: Calculate Moles of H₂ as;

Moles = Mass / M.Mass

Moles = 0.0584 g / 2.01 g/mol

Moles = 0.0290 mol of H₂

Step 2: Find out moles of FeSO₄ as;

According to balance chemical equation,

1 mole of H₂ is produced along with = 1 mole of FeSO₄

So,

0.0290 moles of H₂ will be produced along with = X moles of FeSO₄

Solving for X,

X = 0.0290 × 1 mol / 1 mol

X = 0.0290 moles of FeSO₄

Step 3: Calculate mass of FeSO₄ as;

Mass = Moles × M.Mass

Mass = 0.0290 mol × 151.90 g/mol

Mass = 4.415 g of FeSO₄

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3 years ago

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lora16 [44]

Answer -C They are defined by the number of electrons.
An orbital is a function that describes the probability of finding the electron with certain energy at certain distance from the nucleus therefore, it is defined by the number of electrons.

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Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq) ∆ HbCO(aq) + O2(aq) a. Use

gayaneshka [121]

The question is incomplete, complete question is;

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction:

$HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)$

Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.

$Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8$

$Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306$

Answer:

The equilibrium constant for the given reaction is 170.

Explanation:

$Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8$

$K_1=\frac{[HbO_2]}{[Hb][O_2]}$

$[Hb]=\frac{[HbO_2]}{[K_1][O_2]}$ ..[1]

$Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306$

$K_2=\frac{[HbCO]}{[Hb][CO]}$ ..[2]

$HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)$

$K_c=\frac{[HbCO][O_2]}{[HbO_2][CO]}$ ..[3]

Using [1] in [2]:

$K_2=\frac{[HbCO]}{\frac{[HbO_2]}{[K_1][O_2]}\times [CO]}$

$K_2=K_1\times \frac{[HbCO][O_2]}{[HbO_2][CO]}$

$K_2=K_1\times K_c$ ( using [3])

$306=1.8\times K_c$

$K_c=\frac{306}{1.8}=170$

The equilibrium constant for the given reaction is 170.

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3 years ago