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3 years ago

How many grams of iron are needed to produce 3 g of iron(III) chloride?

Chemistry

1 answer:

Ivanshal [37]3 years ago

4 0

Answer:

5 its just what i do

Explanation:

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Common properties of sodium , calcium , iron ,and gold

Fittoniya [83]

Sodium, calcium, iron and gold are all metals. They all form metallic bonds, in the form of a shared electron cloud. They all have the electrical conductivity and metallic luster which comes with metallic bonds.

7 0

3 years ago

Chlorine gas is added to a large flask to a pressure of 1.85 atm, at a temperature of 322 K. Phosphorus is added, and a reaction

AVprozaik [17]

Answer:

The volume of the flask is<u> 20.245 litres .</u>

Explanation:

We are given with following information-

$PV=nRT$ -------- 1

where R = $0.0821L.atom/mole.K$

Molar mass of $PCl_5=208.22g/mole$

The given chemical equation is -

$2P _(_s_)+5Cl_2 _(_g_)\rightarrow2PCl_5 _(_s_)$ --------- 2

Now , calculation -

Mass of $PCl_5$ formed = 118g

Molar mass of $PCl_5$ = $208.22g/mole$

Mole = $\frac{mass (g)}{molar mass}$

Therefore , moles of $PCl_5$ formed = $\frac{118}{208.22}$

From equation 2 , we get to know that ,

2mole $PCl_5$ formed from 5 mole $Cl_2 _(_g_)$

Therefore , $\frac{118}{208.22}$ mole $PCl_5$ formed from $\frac{5}{2} \times\frac{118}{208.22}$ mole $Cl_2 _(_g_)$

Moles of $Cl_2 _(_g_)$ used = $\frac{5\times118}{2\times208.22} mole$

R= $0.0821L.atom/mole.K$

Pressure (P)= 1.85atm

Temperature (T)= 322K

Moles of $Cl_2 _(_g_)$ (n)= $\frac{5}{2} \times\frac{118}{208.22}$ moles

Applying the formula above in 1 equation , that is

PV = nRT

putting the given values -

$1.85 \times V=\frac{5}{2} \times\frac{118}{208.22}\times0.0821\times322$

V = 20.245 litres.

Hence , the volume of the flask is <u>20.245 litres . </u>

8 0

3 years ago

Write an equation that gives the relationship between the cross-sectional area (a), the volume (v), and the thickness of a cylin

Mazyrski [523]

V = at
Where t is thickness

7 0

1 year ago

C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion please show work

irakobra [83]

Answer:

$\Delta H_{comb}=2043.85kJ/mol$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

$\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}$

Thus, given the values of the enthalpies of formation on the attached file, we obtain: $-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol$

Best regards!

5 0

2 years ago

Thisss for you blake :)))

Lady_Fox [76]

I’m not Blake but ok

4 0

3 years ago

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