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1 year ago

Which chemical or physical change is a endothermic process

Chemistry

2 answers:

Iteru [2.4K]1 year ago

6 0

Answer:

Examples of physical endothermic processes: The melting of ice, the evaporation of water.

PLS MARK ME BRAINLIEST

Explanation:

Ray Of Light [21]1 year ago

5 0

Melting, vaporization, boiling, and sublimation.

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Vladimir [108]

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2 years ago

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Solid NH4HS is introduced into an evacuated flask at 24 ∘C. The following reaction takes place: NH4HS(s)⇌NH3(g)+H2S(g) At equili

Ivanshal [37]

Answer:

0.09425

Explanation:

The reactant is in solid phase and therefore has zero partial pressure.

The products have the same mole ratio (1:1) and will have the same partial pressure = 1/2 × 0.614 atm = 0.307 atm

Kp = (NH3)(H2S) = 0.09425

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2 years ago

One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will

saw5 [17]

Explanation:

The given data is as follows.

Boiling point of water ( $T^{o}_{b}) = 100^{o}C$ = (100 + 273) K = 323 K,

Boiling point of solution ( $T_{b}) = 101.24^{o}C$ = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

$\Delta T_{b} = (T_{b} - T^{o}_{b})$

= 374.24 K - 323 K

= 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

Let molar mass of the solute is x grams.

Therefore, Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

m = $\frac{288 g \times 1000}{x g \times 90}$

= $\frac{3200}{x}$

As, $\Delta T_{b} = k_{b} \times molality$

$1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}$

x = $\frac{0.512 ^{o}C/m \times 3200}{1.24}$

= 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is $C_{n}H_{2n}O_{n}$ .

As, its empirical formula is $CH_{2}O$ and mass is 30 g/mol. Hence, calculate the value of n as follows.

n = $\frac{\text{Molecular mass}}{\text{Empirical mass}}$

= $\frac{1321.29 g}{30 g/mol}$

= 44 mol

Thus, we can conclude that the formula of given material is $C_{44}H_{88}O_{44}$ .

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