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3 years ago

Select the correct answer from each drop-down menu. At chemical equilibrium, the amount of because .

Chemistry

1 answer:

Elena-2011 [213]3 years ago

3 0

Answer:

The answer that completes the question are in BOLD:

At chemical equilibrium, the amount of PRODUCT AND REACTANT REMAIN CONSTANT because the RATES OF THE FORWARD AND REVERSE REACTIONS ARE EQUAL.

Explanation:

In a reversible chemical reaction, an equilibrium is said to be achieved when the rates of the forward reaction is equal to that of the reverse reaction. A reversible reaction is one in which products are formed from reactants simultaneously with the formation of reactants from products.

The combination of two or more substances called REACTANTS gives rise to another substance called PRODUCT, which can in turn give rise to Reactants again. With time, the rate at which the reactants give rise the products, which is called the FORWARD REACTION will be equal to the rate at which the products give rise to the reactants, which is called REVERSE REACTION. At this point, the chemical reaction is said to be in a STATE OF EQUILIBRIUM.

When the rate at which both reaction occurs becomes equal i.e. at an equilibrium state, the concentration of both the reactants and the products becomes constant i.e. no longer changes. Hence, the amount of the reactants forming the products is the same as the amount of products forming the reactants.

N.B: At chemical equilibrium, the amount of the reactants and products does not necessarily equals zero (0). It simply means that there is no net change in the concentration/amount of both reactants and products.

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A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

Nitella [24]

Answer : The specific heat of metal is $0.481J/g^oC$ .

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of metal = 129.00 g

$m_2$ = mass of water = 45.00 g

$T_f$ = final temperature = $39.6^oC$

$T_1$ = initial temperature of metal = $97.8^oC$

$T_2$ = initial temperature of water = $20.4^oC$

Now put all the given values in the above formula, we get

$129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC$

$c_1=0.481J/g^oC$

Therefore, the specific heat of metal is $0.481J/g^oC$ .

4 0

3 years ago

In a chemical process, three bottles of a standard fluid are emptied into a larger container. A study of the individual bottles

Setler79 [48]

Answer:

The expected value is = 39 units

The standard deviation is = 0.1212 unites

The probability is = 0.2047 units

Explanation:

The explanation of this answer is shown on the first uploaded image

5 0

3 years ago

During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.

it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of CH₄ needs → 2 mol of O₂

??? mol of CH₄ needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over * molar mass

= 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0

2 years ago

You decide to clean the bathroom. You notice that the shower is covered in a strange green slime . you try to get rid of this sl

Free_Kalibri [48]

The independent variable (IV) is the lemon juice mixture
The dependent variable (DV) is the appearance of the green slime on the shower
The control variable (CV) are time taken to spray, the amount of spray
The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
The control group (CG) is the side of the shower sprayed with water.

INDEPENDENT VARIABLE

Independent variable is the variable of an experiment that is changed by the experimenter in order to bring about a change. It is the variable being tested in the experiment. In this case, the IV is the lemon juice mixture tested on the green slime on the shower.

DEPENDENT VARIABLE:

Dependent variable is the variable that is observed or measured in an experiment. It is also called responding variable. The DV in this case is the appearance of the green slime on the shower.

CONTROL VARIABLE:

Control variable is the variable that is kept constant throughout the experiment for all groups. The CV is the same for all the groups and they include: time taken to spray, the same amount of spray

CONTROL GROUP

Control group is the group that does not receive the independent variable or test in an experiment. In this case, the CG is the side of the shower sprayed with water.

EXPERIMENTAL GROUP:

Experimental group is the group of ab experiment that receives the experimental treatment or independent variable. In this case, the EG is the side of the shower sprayed with lemon juice mixture.

Therefore, the IV, DV, CV, EG and CG of this experiment are as follows:

The independent variable (IV) is the lemon juice mixture
The dependent variable (DV) is the appearance of the green slime on the shower
The control variable (CV) are time taken to spray, the amount of spray
The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
The control group (CG) is the side of the shower sprayed with water.

Learn more: brainly.com/question/17498238?referrer=searchResults

7 0

2 years ago

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a flas

valentina_108 [34]

Answer:

Kp = 0.022

Explanation:

<em>Full question: ...With 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. </em>

<em />

The equilibrium of ammonia occurs as follows:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Where Kp is defined as:

$Kp = \frac{P_{N_2}P_{H_2}^3}{P_{NH_3}^2}$

<em>Where P represents partial pressure of each gas.</em>

<em />

As initial pressure of ammonia is 2.3atm, its equilibrium concentration will be:

P(NH₃) = 2.3atm - 2X

<em>Where X represents reaction coordinate</em>

<em />

Thus, pressure of hydrogen and nitrogen is:

P(N₂) = X

P(H₂) = 3X.

As partial pressure of hydrogen is 0.69atm:

3X = 0.69

X = 0.23atm:

P(NH₃) = 2.3atm - 2(0.23atm) = 1.84atm

P(N₂) = 0.23atm

P(H₂) = 0.69atm

$Kp = \frac{0.23atm*0.69atm^3}{1.84atm^2}$

8 0

3 years ago