All categories

2 years ago

What phases are most common at high temperatures

Chemistry

1 answer:

Nikitich [7]2 years ago

5 0

Answer:

Liquids and gases.

For most substances, when heated to a high enough temperature, the molecules begin moving faster which results in its liquid or gas state.

You might be interested in

Where is energy stored in living sytems

EastWind [94]

Answer:

energy is stored in the chemical bonds in molecules

Explanation:

6 0

2 years ago

if the climate warmed which of the following would be caused by rising temperatures? .a. an increase in sea levels. b. an increa

Klio2033 [76]

A warming climate can cause seawater to expand and ice over land to melt, both of which can cause a rise in sea level. Storm surge on a Louisiana highway shows the effects of rising sea levels. Many people are interested in climate change and how a changing climate will affect the ocean.

5 0

2 years ago

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

2 years ago

What is the relationship between concentration and rate of reaction?

grigory [225]

<h3><u>Answer;</u></h3>

Directly proportional

<h3><u>Explanation;</u></h3>

<em><u>Concentration is one of the factors that determine the rate of a reaction. Reaction rates increases with increase in the concentration of the reactants, which means they are directly proportional.</u></em>
An increase in the concentration of reactants produces more collisions and thus increasing the rate at which the reaction is taking place. Therefore, <u>Increasing the concentration of a reactant increases the frequency of collisions between reactants and will cause an increase in the rate of reaction.</u>

7 0

3 years ago

How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of th

shutvik [7]

Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

$AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}$

If pH is increased then concentration of $H^{+}$ increases in solution resulting removal of $CH_{3}COO^{-}$ by forming $CH_{3}COOH$ . Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more $AgCH_{3}COO$ will dissolve
If $AgNO_{3}$ is added then concentration of $Ag^{+}$ increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less $AgCH_{3}COO$ will dissolve
Insoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of $Ag^{+}$ . So, more $AgCH_{3}COO$ will dissolve

Hence all three statements are true

8 0

3 years ago