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2 years ago

Where are metals on the table?

Chemistry

1 answer:

lina2011 [118]2 years ago

5 0

21,22,23,24,25,26,27,28,29,39, on the table

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Find the number of Li atoms in 1.50 mole of Li

ale4655 [162]

The molar mass of Li->7g/mol
If 1mol of Li is 7g/mol
1.50mol of Li would be 10.5g/mol

7 0

3 years ago

At 350°c, keq = 1.67 × 10-2 for the reversible reaction 2hi (g) ⇌ h2 (g) + i2 (g). what is the concentration of hi at equilibriu

mariarad [96]

According to the reversible reaction equation:

2Hi(g) ↔ H2(g) + i2(g)

and when Keq is the concentration of the products / the concentration of the reactants.

Keq = [H2][i2]/[Hi]^2

when we have Keq = 1.67 x 10^-2

[H2] = 2.44 x 10^-3

[i2] = 7.18 x 10^-5

so, by substitution:

1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2

∴[Hi] = 0.0033 M

7 0

3 years ago

Scenario Two:

galben [10]

Answer:

jo mom cause why not

Explanation:

7 0

3 years ago

NaHCO3 (s) + HC2H3O2 (aq) = NaC2H3O2 (aq) + H2O (I) + CO2 (g)

Misha Larkins [42]

Moles=volume*concentration
=0.1*.83
=.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2

Mass =Moles*Rfm of CO2
=.083*(12+16+16)
=3.7grams

8 0

3 years ago

Consider the reaction CH4(g) 2O2(g)CO2(g) 2H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for th

zvonat [6]

Answer:

the entropy change for the surroundings when 1.62 moles of CH4(g) react at standard conditions is −8.343 J/K

Explanation:

The balanced chemical equation of the reaction in the question given is:

$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2 H_2O _{(g)}$

Using standard thermodynamic data at 298K.

The entropy of each compound above are listed as follows in a respective order.

Entropy of (CH4(g)) = 186.264 J/mol.K

Entropy of (O2(g)) = 205.138 J/mol.K

Entropy of (CO2(g)) = 213.74 J/mol.K

Entropy of (H2O(g)) = 188.825 J/mol.K

The change in Entropy (S) of the reaction is therefore calculated as follows:

$=1*S(CO2(g)) + 2*S(H2O(g)) - 1*S( CH4(g)) - 2*S(O2(g))$

$=1*(213.74) + 2*(188.825) - 1*(186.264) - 2*(205.138)$

= -5.15 J/mol.K

Given that :

the number of moles = 1.62 of CH4(g) react at standard conditions.

Then;

The change in entropy of the rxn $= 1.62 \ mol * -5.15 \ J/mol.K$

= −8.343 J/K

6 0

3 years ago