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3 years ago

PLEASEEEE HELP :')

Chemistry

1 answer:

mel-nik [20]3 years ago

7 0

Answer:

340 grams Ca₃P₂ (2 sig. figs.)

Explanation:

3Ca + 2P => Ca₃P₂

5.6 mole + excess => ? grams

Convert the 'known' to a coefficient of 1 by dividing all coefficients by 3.

=> Ca + 2/3P => 1/3Ca₃P₂

From the above, 1 mole of Ca => 1/3 mole Ca₃P₂

∴ 5.6 mole Ca in an excess of P => 1/3(5.6 mole) Ca₃P₂

=> 1.8666 mol Ca₃P₂ (calculator answer) ≅ 1.9 mol Ca₃P₂

=> 1.9 mole x 182 g Ca₃P₂/mol Ca₃P₂ = 339.73333 grams Ca₃P₂

≅ 340 grams Ca₃P₂ (2 sig. figs.)

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The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

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Answer:

No the substance is not water.

Explanation:

The balance chemical equation for the decomposition of water is as follow;

2 H₂O = 2 H₂ + O₂

Step 1: Calculate moles of H₂O;

Moles = Mass / M.Mass

Moles = 5.0 g / 18.01 g/mol

Moles = 0.277 moles of H₂O

Step 2: Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:

According to equation,

2 moles of H₂O produced = 1 mole of O₂

So,

0.277 moles of H₂O will produce = X moles of O₂

Solving for X,

X = 0.277 mol × 1 mol / 2 mol

X = 0.138 moles of O₂

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According to equation,

2 moles of H₂O produced = 2 mole of H₂

So,

0.277 moles of H₂O will produce = X moles of H₂

Solving for X,

X = 0.277 mol × 2 mol / 2 mol

X = 0.227 moles of H₂

Step 3: Calculate Mass of O₂ and H₂ as;

For O₂:

Mass = Moles × M.Mass

Mass = 0.138 mol × 31.99 g/mol

Mass = 4.44 g of O₂

For H₂:

Mass = Moles × M.Mass

Mass = 0.227 mol × 2.01 g/mol

Mass = 0.559 g of H₂

Conclusion:

From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

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Answer:

The reactions free energy $\Delta G = -49.36 kJ$

Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

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The pressure of nitrosly chloride (NOCl) is $P_{(NOCl)} = 7.70 \ atm$

The reaction is

$2NO_{(g)} + Cl_2 (g)$ ⇆ $2 NOCl_{(g)}$

From the reaction we can mathematically evaluate the $\Delta G^o$ (Standard state free energy ) as

$\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}$

The Standard state free energy for NO is constant with a value

$\Delta G^o _{NO} = 86.55 kJ/mol$

The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

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$Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }$

Substituting values

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The free energy for this reaction is evaluated as

$\Delta G = \Delta G^o + RT ln Q$

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

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Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

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