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4 years ago

4HF(g)+SiO2(s)→SiF4(g)+2H2O(l)

Chemistry

1 answer:

Alecsey [184]4 years ago

8 0

<span>Gallium-72 that is what i got when i did the math for it</span>

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You have 3.00 m3 of a fixed mass of a gas at 150 kPa. Calculate the pressure if the volume is reduced to 1.20 m3 at a constant t

OverLord2011 [107]

Answer:

375kPa

Explanation:

Pressure * Volume = Constant

150kPa *3m3 =constant

450kPam3=constant

Pressure=constant/volume

450kPam3/1.2m3=375kPa

8 0

3 years ago

A magician pulls a tabecloth out from under dishes and

Kazeer [188]

And he calls it a magic trick even though it is not!

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4 years ago

Which scenario involves a

olga_2 [115]

Answer:

Amount of products and reactants is constant...Although,if any of the factors that interfere this scenerio is involved then there will be a shift or change...

6 0

3 years ago

I need help bad bro i have a 30 cs this is missing

Nat2105 [25]

Answer: 11.02 moles of Ag can be formed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Cu=\frac{350.00g}{63.55g/mol}=5.51moles$

The balanced chemical equation is:

$Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag$

According to stoichiometry :

1 moles of $Cu$ produces = 2 moles of $Ag$

Thus 5.51 moles of $Cu$ will produce= $\frac{2}{1}\times 5.51=11.02moles$ of $Ag$

Thus 11.02 moles of Ag can be formed.

4 0

3 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

3 years ago