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2 years ago

What is the formula for the compound iodine trichoride?

Chemistry

1 answer:

Neko [114]2 years ago

4 0

Answer:

Cl 3 I

Explanation:

Molecular Weight. 233.36. Appearance. Red-orange to brown powder or chunks. Melting Point. 63 °C

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Andrej [43]

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3 years ago

A balloon is filled with 35.0 L of helium in the morning when the temperature is 35.00 oC.

nexus9112 [7]

Answer: V = 33.9 L

Explanation: We will use Charles Law to solve for the new volume.

Charles Law is expressed in the following formula. Temperatures must be converted in Kelvin.

V1 / T1 = V2 / T2 then derive for V2

V2 = V1 T2 / T1

= 35 L ( 308 K ) / 318 K

= 33.9 L

5 0

3 years ago

What is the expected value for the heat of sublimation of acetic acid if its heat of fusion is 10.8 kJ/mol and its heat of vapor

Dennis_Churaev [7]

Answer:

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

Explanation:

$CH_3COOH(l)\rightarrow CH_3COOH(g)$ ..[1]

Heat of vaporization of acetic acid = $H^o_{vap}=24.3 kJ/mol$

$CH_3COOH(s)\rightarrow CH_3COOH(l)$ ..[2]

Heat of fusion of acetic acid = $H^o_{fus}=10.8 kJ/mol$

Heat of sublimation of acetic acid = $H^o_{sub}=?$

$CH_3COOH(s)\rightarrow CH_3COOH(g)$ ..[3]

[1] + [2] = [3] (Hess's law)

$H^o_{sub}=H^o_{vap}+H^o_{fus}$

$=24.3 kJ/mol+10.8 kJ/mol=35.1 kJ/mol$

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

5 0

3 years ago

What is the change in temperature when 75 grams of water releases -2657 J of energy? The specific heat of water is 4.18 J/g°C

Naily [24]

I think the answer for this is 4702.5 J/g*k Depending on if it is water as a solid liquid or gas. I used water as a liquid when I solved it. J=(75g)(4.18 J/g*k)(15K)

8 0

3 years ago

In going from room temperature (25 C) to 10 C above room temperature, the rate of reaction doubles. Calculate the activation ene

Sauron [17]

Answer:

Ea=5.29 × 10⁴ J/mol

Explanation:

In going from 25 °C (298 K) to 35 °C (308 K), the rate of the reaction doubles. Since the rate of the reaction depends on the rate constant (k), this implies that the rate constant doubles. We can find the activation energy (Ea) using the two-point form of the Arrhenius equation.

$ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln\frac{2k_{1}}{k_{1}}=\frac{-Ea}{8.314J/K.mol}.(\frac{1}{308K}-\frac{1}{298K} )\\Ea=5.29 \times 10^{4} J/mol$

8 0

3 years ago