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4 years ago

In the reaction 3H2 + N2 → ____ NH3, what coefficient should be placed in front of NH3 to balance the reaction?

1 answer:

5 0

2. 3H2 + N2 means you have 6 Hs and 2 Ns. NH3 has one N and 3 Hs, so you need 2 NH3s in order to have the 2 and 6 of each that you need on both sides of the reaction.

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You measure the mass of a model car to be 230 grams.The actual mass is 218 grams .what is your percent error

LenKa [72]

Your percent is 5.50458716

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4 years ago

Which two formulas represent compounds

SVEN [57.7K]

CO2 ; H20- They are the only ones that, on both sides, combined with another element and bonding of atoms

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3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

meriva

Answer:

The standard enthalpy change for the reaction at $25^{0}\textrm{C}$ is -2043.999kJ

Explanation:

Standard enthalpy change ( $\Delta H_{rxn}^{0}$ ) for the given reaction is expressed as:

$\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

Where $\Delta H_{f}^{0}$ refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

$\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ$

4 0

3 years ago

Atomic number of a Carbon atom is 6. How many electrons does it have?

Nostrana [21]

In the first shell there is 2 electrons and on the second shell there is 4 elections.

6 0

3 years ago