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3 years ago

Calculate how many grams of each product are formed when 4.05g of h2o is used. 2h2o=2h2+o2

Chemistry

1 answer:

NISA [10]3 years ago

3 0

Answer:

Explanation:

Given data:

Mass of water used = 4.05 g

Mass of each product produced = ?

Solution:

Chemical equation:

2H₂O → 2H₂ + O₂

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 4.05 g/ 18 g/mol

Number of moles = 0.225 mol

Now we will compare the moles of water with hydrogen and oxygen.

H₂O : H₂

2 : 2

0.225 : 0.225

H₂O : O₂

2 : 1

0.225 : 1/2×0.225 = 0.113 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 0.225 × 2 g/mol

Mass = 0.45 g

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.113 mol × 32 g/mol

Mass = 3.616 g

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Explanation:

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3 years ago

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o

Damm [24]

Answer : The partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of $HBr$ = $1.0\times 10^{-2}atm$

The partial pressure of $H_2$ = $2.0\times 10^{-4}atm$

The partial pressure of $Br_2$ = $2.0\times 10^{-4}atm$

$K_p=4.2\times 10^{-9}$

The balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴

At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}$

Now put all the values in this expression, we get :

$4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}$

$p=-1.99\times 10^{-4}$

The partial pressure of $H_2$ at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

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4 years ago

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Answer:

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Explanation:

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3 years ago

Read 2 more answers

At 35.0°c and 3.00 atm pressure, a gas has a volume of 1.40 l. what pressure does the gas have at 0.00°c and a volume of 0.950 l

Leona [35]

Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

Solution :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 3 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 1.40 L

$V_2$ = final volume of gas = 0.950 L

$T_1$ = initial temperature of gas = $35^oC=273+35=308K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}$

$P_2=3.918atm$

Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

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3 years ago

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What type(s) of intermolecular forces are expected between ch3ch2ch2ch2ch2oh molecules?

notka56 [123]

Answer : Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.

Explanation :

The given molecule is, $CH_3CH_2CH_2CH_2CH_2OH$

Three types of inter-molecular forces are present in this molecule which are Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.

Hydrogen-bonding : when the partial positive end of hydrogen is bonded with the partial negative end of another molecule like, oxygen, nitrogen, etc.

Dipole-dipole attraction : When the partial positively charged part of the molecule is interact with the partial negatively charged part of the molecule. For example : In case of HCl.

London-dispersion force : This force is present in all type of molecule whether it is a polar or non-polar, ionic or covalent. For example : In case of Br-Br , F-F, etc

Hydrogen-bonding is present between the oxygen and hydrogen molecule.

Dipole-dipole forces is present between the carbon and oxygen molecule.

London-dispersion forces is present between the carbon and carbon molecule.

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4 years ago