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3 years ago

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How do you solve this? I am so confused. This is Chemistry 1 HON and its dimensional analysis. It would help me so much if someo

ne explained it and showed work please and thank you...

1 answer:

Mekhanik [1.2K]3 years ago

4 0

1. 150 yards

2. 4320

3. 144

4. 2688

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How is chemical bonding useful to our lives?

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Answer:

Three types of chemical bonds are important in human physiology, because they hold together substances that are used by the body for critical aspects of homeostasis, signaling, and energy production, to name just a few important processes. These are ionic bonds, covalent bonds, and hydrogen bonds.

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2 years ago

A calorie is another term for carbohydrate.TF

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2 years ago

When 50.0 mL of 1.27 M of HCl(aq) is combined with 50.0 mL of 1.32 M of NaOH(aq) in a coffee-cup calorimeter, the temperature of

sergij07 [2.7K]

Answer:

-55.9kJ/mol is the change in enthalpy of the reaction

Explanation:

In the reaction:

HCl(aq) + NaOH(aq) → H₂O(l) + NaCl

Some heat is released per mole of reaction.

To know how many moles reacts we need to find limiting reactant:

Moles HCl = 0.050L ₓ (1.27mol / L) = 0.0635 moles HCl

Moles NaOH = 0.050L ₓ (1.32mol / L) = 0.066 moles NaOH

As there are more moles of NaOH than moles of HCl, <em>HCl is limiting reactant and moles of reaction are moles of limiting reactant, </em><em>0.0635 moles</em>

<em />

Using the coffee-cup calorimeter equation we can find how many heat was released thus:

Q = C×m×ΔT

<em>Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)</em>

Replacing:

Q = 4.18J/g°C×100g×8.49°C

Q = 3548.8J of heat are released in the reaction

Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:

ΔH = Heat / mol of reaction

ΔH = -3548.8J / 0.0635 moles of reaction

<em>Negative because is released heat. </em>

ΔH = -55887J / mol

ΔH =

<h3>-55.9kJ/mol is the change in enthalpy of the reaction</h3>

<em />

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2 years ago

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3 years ago

Read 2 more answers

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = $\frac{20}{125}$ = 0.16M

<u>Molarity of H20</u>:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

4 0

3 years ago