What 2 types of rock are most common beneath Earth’s crust?*

A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h

NikAS [45]

Nitrogen combine with hydrogen to produce ammonia $\text{NH}_3$ at a $1:3:2$ ratio:

$\text{N}_2 \; (g) + 3 \; \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)$

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. $3 \; \text{mol}$ of hydrogen gas would have been consumed while $2 \; \text{mol}$ of ammonia would have been produced. The final mixture would therefore contain

$17 \; \text{mol}$ of $\text{H}_2 \; (g)$ and
$2 \; \text{mol}$ of $\text{NH}_3 \; (g)$

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

$\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.926 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.723 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 2.037 \times 10^{2} \; \text{kPa} \end{array}$

6 0

3 years ago

in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co

kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

$\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5$

mol of element G :

$\tt mol=\dfrac{4}{16}=0.25$

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0

3 years ago

Rank these compounds by their expected boiling point? CH4 CH3Cl CH3OH? Lowest to highest; 10+ for the best explanation. Thank yo

docker41 [41]

Methane, CH4, would have the lowest boiling point among the three since it has the lowest number of carbon and has no functional groups. Methanol would have the highest boiling point since it has a functional group which contains hydrogen bonding which much stronger than the one in CH3Cl. Hope this helps.<span />

3 0

3 years ago

Read 2 more answers

Is ineffective collision's the answer? Please correct me if not, thank you. :)

AleksAgata [21]

Yes! You're correct! Hope this helps! :D

6 0

3 years ago

Question 7 unsaved liquid water can store more heat energy than an equal amount of any other naturally occurring substance becau

Paul [167]

Has the highest specific heat.
___________

3 0

3 years ago