All categories

3 years ago

If you have 3.54 g of H2, how many grams of NH3 can be produced?

1 answer:

5 0

3.54g H2 means you have 3.54 / 2 = 1.77 moles of H2, which means you have 1.77*2 = 3.54 moles of H atom.

Since 1 mole of NH3 has 3 moles of H, so you can produce 3.54 / 3 moles of NH3, which is (3.54/3) * (12+3) = 17.7 grams

You might be interested in

Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is

Citrus2011 [14]

Depression in freezing point (Δ $T_{f}$ ) = $K_{f}$ ×m×i,
where, $K_{f}$ = cryoscopic constant = $1.86^{0} C/m$ ,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For $NaNO_{3}$ )

Thus, (Δ $T_{f}$ ) = 1.86 X 0.0085 X 2 = $0.03162^{0}C$

Now, (Δ $T_{f}$ ) = $T^{0}$ - T
Here, T = freezing point of solution
$T^{0}$ = freezing point of solvent = $0^{0}C$
Thus, T = $T^{0}$ - (Δ $T_{f}$ ) = - $0.03162^{0}C$

8 0

3 years ago

Read 2 more answers

Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o

Nata [24]

<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M$

where,

$n_p$ = number of protons

$m_p$ = mass of one proton

$n_n$ = number of neutrons

$m_n$ = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is $_{15}^{31}\textrm{P}$

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

$\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399$

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

8 0

2 years ago

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a

Inessa [10]

Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits is mathematically given as

N=1820slits

<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>

Generally, the equation for the wave is mathematically given as

$d\ sin\ (\theta\ m) \ = \ m\ \lambda$