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3 years ago

Consider the chemical equation in equilibrium.

Chemistry

2 answers:

Tema [17]3 years ago

8 0

Answer: A. The equilibrium will shift to the left to favor the reverse reaction.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$CH_4(g)+H_2O(g)\leftrightharpoons CO(g)+3H_2$

If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatelier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction to favor the formation of reactants in reverse reaction.

Kruka [31]3 years ago

4 0

By increasing the P reaction moves toward less number of moles in this case
<span>The equilibrium will shift to the left to favor the reverse reaction.</span>

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a chemist produces 460 mL of oxygen gas at -43 C and constant pressure. to what celsius temperature must the oxygen be warmed in

Fudgin [204]

Answer:

27 °C

Explanation:

Applying,

V/T = V'/T'................. Equation 2

Where V = Initial volume of oxygen, T = Initial temperature, V' = Final volume of oxygen, T' = Final temperature.

Make T' the subject of the equation

T' = V'T/V................ Equation 2

Form the question,

Given: V' = 600 mL, V = 460 mL, T = -43°C = (-43+273) = 230K

Substitute these values into equation 2

T' = (600×230)/460

T' = 300 K

T' = (300-273) °C

T' = 27 °C

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3 years ago

xxMikexx [17]

Answer:

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3 years ago

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45

musickatia [10]

Answer:

For 1 antacid tablet (in ballon1) we get .0173 moles of CO2

for 2 tablets (in balloon 2) we get: 2*0,0173= 0.0346 moles of CO2

For 3 tablets (in balloon 3) we get 3* 0.0173 = 0.0519 moles of CO2

Explanation:

The complete question:

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon).

Balloon 1 had 1 antacid tab

Baloon 2 had 2

Balloon 3 had 3

Step 1: Data given

1.45 g of NaHCO3 is present in an antacid tablet

Molar mass of NaHCO3 = 84.00 g/mol

Step 2: The balanced equation

NaHCO3 + H2O → NaOH + H2O + CO2

Step 3: Calculate moles of NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

1.45g / 84.0 g/mol = .0173 moles

Step 4: Calculate moles CO2

For 1 mol NaHCO3 we need 1 mol H2O to produce 1 mol NaOH 1 mol H2O and 1 mol CO2

For 0.0173 moles NaHCO3 we'll get 0.0173 moles CO2

so for 1 antacid tablet we get .0173 moles of CO2

for 2 tablets we get: 2*0,0173 = 0.0346 moles of CO2

For 3 tablets we get 3* 0.0173 = 0.0519 moles of CO2

5 0

3 years ago

Calculate the amount of heat deposited on the skin of a person burned by 1.00g of steam at 100.0C. Skin temperature is 37C. ΔHva

seraphim [82]

The amount of heat deposited on the skin is 2.26 kJ.

<h3>What is the amount of heat given off by 1.0 g of steam?</h3>

The amount of heat given off by steam is determined using the formula below:

Quantity of heat = mass * latent heat of vaporization.

Moles of steam = 1.0/18

Heat = 1.0/18 * 40.7

Heat deposited = 2.26 kJ

In conclusion, the quantity of heat is determined from the latent heat of vaporization and the moles of steam.

Learn more about heat of vaporization at: brainly.com/question/26306578

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7 0

2 years ago

The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of

V125BC [204]

Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

$C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O$

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

$3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})$

= 1401.5 kJ

Now, let's convert kJ to kcal.

We know that, 1kcal = 4.184kJ

So, $1401.5kJ(\frac{1kcal}{4.184kJ})$

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.

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3 years ago