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Temka [501]
3 years ago
5

How many grams of H2 are needed to produce 10.42 g of NH3

Chemistry
2 answers:
tamaranim1 [39]3 years ago
8 0
N2 + 3 H2 >> 2 NH3 moles NH3 = 11.50 g /17.0307 g/mol=0.6753 the ratio between H2 and NH3 is 3 : 2 moles H2 needed = 0.6753 x 3/2 =1.013 mass H2 = 1.013 mol x 2.106 g/mol=2.042 g
RideAnS [48]3 years ago
8 0

Answer:

1.84g of H₂ are needed to produce 10.42g of NH₃

Explanation:

In the production of ammonia, NH₃, the following reaction occurs

N₂ + 3H₂ ⇄ 2NH₃

From, the above, <em>3 moles</em> of hydrogen gas (H₂) is required to produce <em>2 moles</em> of ammonia (NH₃).

The atomic mass of Hydrogen is <em>1 g/mol</em>, <em>3 moles</em> of hydrogen gas are present in the reactant, hence the mass of hydrogen gas in the equation is;

3 × 2 × 1 = <u>6g</u>

The atomic mass of Nitrogen is <em>14g/mol</em>, that of hydrogen is <em>1g/mol</em> and <em>2 moles</em> of ammonia were produced. Hence the mass of ammonia in the equation is;

(2 × 14) + (2 × 3 × 1) = <u>34g</u>

<u />

From the above, it can be deduced that 6g of H₂ is required to produce 34g of NH₃, hence "x" grams of H₂ are needed to produce 10.42g of NH₃

To find x (which is the unknown)

6g ⇒ 34g

x ⇒ 10.42g

x = \frac{10.42 * 6}{34}

x = 1.84g

1.84g of H₂ are needed to produce 10.42g of NH₃

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Answer:

Acceleration

Explanation:

The rate at which velocity changes with time is called Acceleration is a measure of how quickly the velocity is changing. If velocity does not change, there is no acceleration.

5 0
3 years ago
A reaction between substances Y and Z is
Anuta_ua [19.1K]

Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

K=A\times e^{\frac{-Ea}{RT}}

Taking logarithm on both the sides, we get:

\ln k=-\frac{Ea}{RT}+\ln A             ............(1)

where,

k = rate constant  = 2.95\times 10^{-3}L/mol.s

Ea = activation energy  = ?

T = temperature = 435 K

R = gas constant  = 8.314 J/K.mole

A = pre-exponential factor  = 3.00\times 10^{+10}L/mol.s

Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)

Ea=108318.365J/mol=108.318kJ/mol

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol

3 0
3 years ago
Isotopes of the same element will have a different number of neutrons, but the same number of protons. TRUE OR FALSE
Nikitich [7]

Answer:

TRUE

Explanation:

6 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
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The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

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<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
What is the mass of a magnesium block that measures 2.00 cm
skad [1K]

Answer:

m = 31.284 grams

Explanation:

Given that,

The dimension of a magnesium block is 2.00 cm  x 3.00 cm x 3.00 cm.

The density of magnesium is, d = 1.738 g/cm³

We need to find the mass of the magnesium block. We know that the density of an object is given by its mass per unit its volume. So,

d=\dfrac{m}{V}\\\\\text{Where m=mass}\\\\m=d\times V\\\\m=1.738\ g/cm^3\times (2\times 3\times 3)\ cm^3\\\\m=31.284\ \text{grams}

So, the mass of the block is 31.284 grams.

5 0
3 years ago
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