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2 years ago

As temperature increase and the volume stays constant, how will the pressure change

Chemistry

1 answer:

crimeas [40]2 years ago

5 0

Answer:

There will be more collisions and so a greater pressure. The number of particles is proportional to pressure, if the volume of the container and the temperature remain constant. ... Volume is inversely proportional to pressure, if the number of particles and the temperature are constant.

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CH4 (g) yields C(g) + 4H (g) (reaction for expansion)

anyanavicka [17]

Answer:

Here's what I find.

Explanation:

$\rm CH$_{4}$(g) $\, \rightleftharpoons \,$ C(g) + 4H(g)$

$\text{Same} =\begin{cases}1. & \text{The number of atoms of each element}\\2. & \text{The state of the methane molecules}\\4. &\text{The state of the carbon atoms}\\\end{cases}$

$\text{Different} =\begin{cases}3. & \text{The enthalpy change of the reaction}\\\end{cases}$

The sign of ΔH changes when you reverse the reaction.

5 0

3 years ago

Read 2 more answers

PLEASE HELP CHEMISTRY

eduard

When sodium carbonate is dissolved in water, the equation is $Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq)$ .

When carbon dioxide is placed in water, aqueous carbon dioxide is formed: $CO_2 (g) --- > CO_2(aq)$

<h3>Dissolution of compounds in water</h3>

Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.

Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.

While in solution, sodium carbonate dissociates into its component ions according to the following equation:

$Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq)$

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.

Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: $CO_2 (g) --- > CO_2(aq)$

More on the dissolution of substances can be found here: brainly.com/question/28580758

#SPJ1

7 0

1 year ago

A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns c

makkiz [27]

0.216g of aluminium compound X react with an excess of water water to produce gas. this gas burn completely in O2 to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure

0.108 / n = 24 / 1
n = 0.0045 mole ( CO2 >>0.0045 mole
0.216 - 0.0045 = 0.2115
so Al = 0.2115 / 27 => 0.0078 mole
C = 0.0045 * 1000 => 4.5 and Al = 0.0078 * 1000 = 7.8

7 0

3 years ago

Which of the following describes the role of water in supporting the chemical reactions that occur within various life forms? It

wel

Answer:

The correct option is;

It is used during photosynthesis to capture sunlight

Explanation:

During photosynthesis, light energy from the Sun is converted and stored in sugars as chemical energy. The Sun light energy is used in the formation of complex sugars such as glucose from the combination of water from the ground and carbon dioxide from the atmosphere while oxygen is released as the byproduct. Organisms are then able to obtain energy from the glucose as well as carbon fiber

The chemical equation for the reaction is as follows;

6CO₂ + 12H₂O + light energy → C₂H₁₂O₆ + 6O₂ + 6H₂O

Carbon, Water, GLucose, Oxygen, Water

dioxide

4 0

3 years ago

Read 2 more answers

Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,

storchak [24]

Answer:

The mass of copper(II) sulfide formed is:

= 81.24 g

Explanation:

The Balanced chemical equation for this reaction is :

$Cu(s) + S\rightarrow CuS$

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{54}{63.55}$

Moles of Cu = 0.8497 mol

Given mass = 42 g

Molar mass of S = 32.06 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{42}{32.06}$

Moles of S = 1.31 mol

Limiting Reagent : The reagent which is present in less amount and consumed in a reaction

First find the limiting reagent :

$Cu + S\rightarrow CuS$

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require = 1 x 0.8497 mol

= 0.8497 mol of S

S present in the reaction Medium = 1.31 mol

S Required = 0.8497 mol

S is present in excess and Cu is limiting reagent

All Cu is consumed in the reaction

Amount Cu will decide the amount of CuS formed

$Cu + S\rightarrow CuS$

1 mole of Cu gives = 1 mole of Copper sulfide

0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$0.8497 = \frac{given\ mass}{95.611}$

Mass of CuS = 0.8497 x 95.611

= 81.24 g

3 0

3 years ago