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ludmilkaskok [199]

3 years ago

5

ice _______________ - a collective term for several mechanical weathering processes induced by stresses created by the freezing

of water into ice.

1 answer:

elena-14-01-66 [18.8K]3 years ago

6 0

Answer: Frost weathering is a collective term for several mechanical weathering processes induced by stresses created by the freezing of water into ice.

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Write a word equation for the combustion of propane???<br> HELP????

frutty [35]

Answer:

1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.

Explanation:

The word equation for the combustion of propane can be obtained from the chemical equation;

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

The word equation is therefore:

1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.

For such a combustion reaction, carbon dioxide and water are produced in the process.

8 0

3 years ago

An amino acid A, isolated from the acid-catalyzed hydrolysis of a peptide antibiotic, gave a positive ninhydrin test and had a s

Ede4ka [16]

The answer could be 23

7 0

3 years ago

Never mind jhvjycdtrsesetdfyguhbjnk

OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

So

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

So

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

So

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

6 0

3 years ago

Read 2 more answers

Which statements are true concerning elements in the same group of the periodic table? Select all that apply.

Tatiana [17]

The statements A and F are true concerning elements in the same group of the periodic table.

To select all that apply, we need to evaluate each statement.

A. They have similar periodic properties.

This statement is true. All the elements in the same group will have similar chemical and physical properties due to the electron configuration of their outer shell. For example, the alkali metals group (Li, Na, K, Rb, Cs, and Fr), has a valence electron configuration of s¹ (in the outer shell), which gives them the tendency to react vigorously with water, as well as other properties.

B. They are all metals or nonmetals, but not both.

This is false. If we take a look at the p-block of the periodic table, we can see that the <u>groups</u> in this block are conformed by nonmetals, metals, and metalloids. For example, the icosagens group is formed by metalloids (B) and metals (Al, Ga, In, Tl).

C. They are either all solids or all liquids or all gases.

This is false. In some groups, all the elements are solids (alkaline earth metals) or gases (group of noble gases), but in others, the groups are conformed by gases with solids (pnictogens group) or by gases with liquids (halogens group).

D. They have the same number of shells of electrons.

This is false. In a group, the number of shells <u>increases from top to bottom</u> in the periodic table. For example, the electron configuration of the elements in the alkali metals is:

H: 1s¹
Li: [He]2s¹
Na: [Ne]3s¹
K: [Ar]4s¹
Rb: [Kr]5s¹
Cs: [Xe]6s¹
Fr: [Rn]7s¹

We can see that hydrogen has 1 shell and Cs has 6 shells.

E. They have the same number of inner core electrons.

This is false. As we said at point D, the number of shells increases from top to bottom in a group, so the number of inner core electrons also increases in this order. For example, in the alkaline earth metals group, the electron configuration of the elements is:

Be: [He]2s²
Mg: [Ne]3s²
Ca: [Ar]4s²
Sr: [Kr]5s²
Ba: [Xe]6s²
Ra: [Rn]7s²

As we can see, the number of inner shells increases from Be ([He]) to Ra ([Rn]).

F. They have the same outer shell electron configuration.

This is true. As we said at point A, the elements in the same group will have the same electron configuration of the outer shell (valence electron configuration). At points D and E, we can see that the valence electron configuration is the same for all the elements in the groups.

Therefore, statements A and F are true.

You can find more about the periodic table here: brainly.com/question/4287157?referrer=searchResults

I hope it helps you!

6 0

3 years ago

I will give brainlyst for this so pls help

omeli [17]

Answer:

Answer -b

Explanation:

Brainliest pls

8 0

3 years ago

Read 2 more answers