What is one important discovery that the curies made

In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 2.8

REY [17]

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3 0

3 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

I need help ASAP!!!!! What happens to water when it changes to ice?

Gre4nikov [31]

- It's density increases

- mass increases

- volume increases

Hope this helps!

6 0

2 years ago

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Determine how many mm of Hg are equal to 5.3 atm of pressure

Leno4ka [110]

4028 mm of Hg...........

5 0

3 years ago

A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa

german

Boyle's law of ideal gas: This law states that the volume of a gas is inversely proportional to its pressure at a constant temperature. Acc to this law we can write the relation of pressure and volume as:

$PV=Constant$

That means:

$P_{1}V_{1}=P_{2}V_{2}$

From that equation we can calculate Volume of gas at a certain pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

Here P₁, initial pressure is given as 85.0 kPa

V₁, initial volume is given as 525 mL

P₂, final pressure is 65.0 kPa

$P_{1}V_{1}=P_{2}V_{2}$

so,

V_{2}=85\times 525\div 65

=686 mL

Volume of gas will be 686 mL.

8 0

3 years ago

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