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IrinaVladis [17]
3 years ago
14

What is the mass % of iron in Fe(CH3CO2)2 (Fe=iron=55.85)

Chemistry
1 answer:
Serjik [45]3 years ago
5 0
The mass percentage is 15.1465%.
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Describe how glacier mice are complete ecosystems, even though they are tiny.<br> GIVING BRAINIEST
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Answer: The clumps, known as glacier mice, have been found to contain miniature ecosystems

Explanation: Because glaciers are in constant, if slow, motion and are frequently blasted by strong winds.

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2 years ago
Nitrogen dioxide is a red-brown gas that is responsible for the color of photochemical smog. What is the volume of 1 mol of nitr
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B ideal gas has a volume of 22.4
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3 years ago
Write the autoionization reaction for methanol, ch3oh.
aliina [53]
Autoionization Reactions are those reactions in which ions or molecules ionizes spontaneously without adding any external reagent.

For Example,
                    Autoionization of water.

                               H₂O  +  H₂O   ⇆   H₃O⁺  +  OH⁻

Autoionization reaction of Methanol is shown below,

4 0
4 years ago
He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
Xelga [282]

Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

8 0
3 years ago
Read 2 more answers
A 20.0g sample of metal with a specific heat of 5 J/(g°C) raised it's initial temperature to 40.0 when 500J heat was added. What
slavikrds [6]

The initial temperature of the metal = 35 °C

<h3>Further explanation</h3>

Heat can be formulated :

Q = m . c . ΔT

Q = heat, J

c = specific heat, J/g C

ΔT = temperature, °C

m = 20 g

c = 5 J/(g°C)

Q = 500 J

T₁ = 40 C

the initial temperature :

\tt \Delta t(T_2-T_1)=\dfrac{Q}{m.c}\\\\40-T_1=\dfrac{500}{20.5}\\\\40-T_1=5\\\\T_1=35^oC

3 0
3 years ago
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