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julia-pushkina [17]

3 years ago

14

Consider the reaction N2(

2 answers:

Tanzania [10]3 years ago

7 0

Answer:

The reaction shifts toward the product gas.

Explanation:

DENIUS [597]3 years ago

3 0

N₂ + 3H₂ ⇆ 2NH₃

↓ volume = ↑ pressure

The reaction shifts toward the product gas.

:)

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Doy CORONITA a quien me ayude rapido que es un examen porfaaa

rodikova [14]

Answer:

thank you for the points

8 0

3 years ago

The density of concentrated ammonia, which is 28.0% w/w nh3, is 0.899 g/ml. what volume of this reagent should be diluted to 1.0

vlada-n [284]

Answer: 2.4 ml

Solution :

Molar mass of $NH_3$ = 17 g/mole

Given,: 28% w/w of $NH_3$ solution means 28 g of ammonia in 100 g of solution.

Mass of solution = 100 g

Now we have to calculate the volume of solution.

$Volume=\frac{Mass}{Density}=\frac{100g}{0.899g/ml}=111.2ml$

Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute $NH_3=\frac{\text {given mass}}{\text {molar mass}}=\frac{28}{17}=1.65moles$

$V_s$ = volume of solution in liter = 0.11 L

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.65moles}{0.11L}=15mole/L$

Using molarity equation:

$M_1V_1=M_2V_2$

$15\times V_1=0.036\times 1.0\times 10^{3}$

$V_1=2.4ml$

6 0

3 years ago

calculate the mols of alt gas if the volume is 0.97 liters at a temperature of 12 C and the pressure is 152 Kpa’s

katrin2010 [14]

Answer:

0.062mol

Explanation:

Using ideal gas law as follows;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821Latm/molK)

T = temperature (K)

Based on the information provided;

P = 152 Kpa = 152/101 = 1.50atm

V = 0.97L

n = ?

T = 12°C = 12 + 273 = 285K

Using PV = nRT

n = PV/RT

n = (1.5 × 0.97) ÷ (0.0821 × 285)

n = 1.455 ÷ 23.39

n = 0.062mol

4 0

3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

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3 years ago

How would you describe the kinetic energy of the particles in a solid? 0 low kinetic energy O high kinetic energy same kinetic e

TEA [102]

Answer:

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2 years ago