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2 years ago

Fe+H2SO4------>????????????

Chemistry

2 answers:

viktelen [127]2 years ago

6 0

Answer:

When iron reacts with dilute H2SO4, iron (II) sulphate with the evolution of hydrogen gas is formed.

yanalaym [24]2 years ago

4 0

<em>equation:</em> Iron + Sulfuric acid → Iron (II) sulfate + hydrogen gas

<em>Type of Chemical Reaction:</em> For this reaction we have a single displacement reaction.

Note that this is the chemical reaction that takes place when diluted Sulfuric acid reacts with iron. If you have concentrated Sulfuric acid you will have a different equation.

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Write the chemical symbol and the charge for the ion with 7 protons and 10 electrons.

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It’s Nitrogen (N) with a -3 charge

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3 years ago

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3 years ago

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If a force of 1.4N is applied to a block with a mass of 7kg, what is the acceleration of the block in meters per second squared?

Anton [14]

Answer:

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$a = \frac{1.4}{7} = \frac{1}{5} \\$

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2 years ago

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What is the conversion of 85.0kg

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4 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago